0.头文件
#define _CRT_SBCURE_NO_DEPRECATE#include <set>#include <cmath>#include <queue>#include <stack>#include <vector>#include <string>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>#include <functional>using namespace std;const int maxn = 110;const int INF = 0x3f3f3f3f;1.埃拉托斯特尼筛法
/* |埃式筛法| |快速筛选素数| |16/11/05ztxint prime[maxn]; bool is_prime[maxn];int sieve(int n){ int p = 0; for(int i = 0; i <= n; ++i) is_prime[i] = true; is_prime[0] = is_prime[1] = false; for (int i = 2; i <= n; ++i){ // 注意数组大小为n if(is_prime[i]){ prime[p++] = i; for(int j = i + i; j <= n; j += i) // 轻剪枝，j必须是i的倍数 is_prime[j] = false; } } return p; // 返回素数的数量2.快速幂
/* |快速幂| |16/11/05ztxtypedef long long LL; // LL取决于视数据的大小 powerMod(LL x, LL n, LL m){ LL res = 1; while (n > 0){ if (n & 1) // 判断是否为奇数，如果是true res = (res * x) % m; x = (x * x) % m; n >>= 1; // 相当于n /= 2; } return res;}3.大数模拟
大数加法
/* |大数模拟加法| |用string模拟| |16/11/05ztx， thanks to caojiji|*/string add1(string s1, string s2){ if (s1 == "" && s2 == "") return "0"; if (s1 == "") return s2; if (s2 == "") return s1; string maxx = s1, minn = s2; if (s1.length() < s2.length()){ maxx = s2; minn = s1; } int a = maxx.length() - 1, b = minn.length() - 1; for (int i = b; i >= 0; --i){ maxx[a--] += minn[i] - '0'; // a一直在减 ， 额外的还要减少‘0’ } for (int i = maxx.length()-1; i > 0;--i){ if (maxx[i] > '9'){ maxx[i] -= 10;//注意这一点，减10 maxx[i - 1]++; } } if (maxx[0] > '9'){ maxx[0] -= 10; maxx = '1' + maxx; } return maxx;}大数阶乘
/* |大数模拟阶乘| |用数组模拟| |16/12/02ztx#include <iostream>#include <cstdio>using namespace std;typedef long long LL;const int maxn = 100010;int num[maxn], len;/* 在mult函数中，形参部分：len每次调用函数都会发生变化，n表示每次乘以的数量，最终返回结果的长度 tip: 阶乘都是先求之前的(n-1)!来求n! Init函数的初始化非常重要，不要落后*/void Init() { len = 1; num[0] = 1;}int mult(int num[], int len, int n) { LL tmp = 0; for(LL i = 0; i < len; ++i) { tmp = tmp + num[i] * n; //从最低位开始，等号左侧的tmp表示当前位置，右侧的tmp表示进位(前进位) num[i] = tmp % 10; // 保存在相应的数组位置，即移除进位后的一位数 tmp = tmp / 10; // 再循环采用取整，与n和下一个位置的乘积相加 } while(tmp) { // 进位处理后 num[len++] = tmp % 10; tmp = tmp / 10; } return len;}int main() { Init(); int n; n = 1977; // 求的阶乘数 for(int i = 2; i <= n; ++i) { len = mult(num, len, i); } for(int i = len - 1; i >= 0; --i) printf("%d",num[i]); // 从最高位置依次输出，printf输出用于更多数据 printf("\n"); return 0;}4.GCD
/* |辗转相除法| |欧几里算法| |最大公约数| |16/11/05ztxint gcd(int big, int small){ if (small > big) swap(big, small); int temp; while (small != 0){ // 辗转相除法 if (small > big) swap(big, small); temp = big % small; big = small; small = temp; } return(big);}5.LCM
/* |辗转相除法| |欧几里算法| |要求最小公倍数| |16/11/05ztxint gcd(int big, int small){ if (small > big) swap(big, small); int temp; while (small != 0){ // 辗转相除法 if (small > big) swap(big, small); temp = big % small; big = small; small = temp; } return(big);}6.全排列
/* |要求1-n的全排列， 有条件| |16/11/05ztx， thanks to wangqiqi|void Pern(int list[], int k, int n) { // k表示前k个数不动，后n-k位数不动 if (k == n - 1) { for (int i = 0; i < n; i++) { printf("%d", list[i]); } printf("\n"); }else { for (int i = k; i < n; i++) { // 所有满足移动条件的全排列输出 swap(list[k], list[i]); Pern(list, k + 1, n); swap(list[k], list[i]); } }}7.二分搜索
/* |二分搜索| |要求：先排序| |16/11/05ztx， thanks to wangxiaocai left是最初的元素， right是末尾元素的下一个数字，x是要找到的数字 bsearch(int *A, int left, int right, int x){ int m; while (left < right){ m = left + (right - left) / 2; if (A[m] >= x) right = m; else left = m + 1; // 若要替换 upper_bound, 改为:if (A[m] <= v) x = m+1; else y = m; } return left;}/* 最后left == right 如果找不到13577找6，返回7 如果有多少x，可以用lower_bound搜索，upper_bound搜索，下标相减 C++自带lower__bound(a,a+n,x)返回数组中最后一个x的下一个数字的地址 upper_bound(a,a+n,x)返回数组中第一个x的地址 如果在a+n中没有找到x或x的下一个地址，请返回a+n的地址 lower_bound(a,a+n,x）-upper_bound(a,a+n,x)返回数组中x的个数*/8.并查集
/* |合并节点操作| |16/11/05ztx， thanks to chaixiaojunint father[maxn]; // father父节点存储i void makeSet() { for (int i = 0; i < maxn; i++) father[i] = i; } int findRoot(int x) { // 迭代找根节点 int root = x; // 根节点 while (root != father[root]) { // 寻找根节点 root = father[root]; } while (x != root) { int tmp = father[x]; father[x] = root; // 根节点赋值 x = tmp; } return root; } void Union(int x, int y) { // 将x所在的集合与y所在的集合结合形成一个集合。 int a, b; a = findRoot(x); b = findRoot(y); father[a] = b; // y连接到x的根节点 或father[b] = a是x连接到y的根节点； } /* findroot(x)中： 路径压缩 迭代 最优版 关键是路径上的每个节点都可以直接连接到根*/最小生成树
Kruskal9.克鲁斯卡尔算法
/* |Kruskal算法| |适用于 稀疏图 要求最小生成树| |16/11/05ztx thanks to wangqiqi|* 第一步：点、边、加vector，把所有的边从小到大排序 第二步:并收集部分 + 下面的code*/void Kruskal() { ans = 0; for (int i = 0; i<len; i++) { if (Find(edge[i].a) != Find(edge[i].b)) { Union(edge[i].a, edge[i].b); ans += edge[i].len; } } }10.普里姆算法
/* |Prim算法| |适用于 稠密图 要求最小生成树| |堆优化版，时间复杂度:O(elgn)| |16/11/05ztx， thanks to chaixiaojunstruct node { int v, len; node(int v = 0, int len = 0) :v(v), len(len) {} bool operator < (const node &a)const { // 加入队列的元素从小到大自动按距离排序 return len> a.len; } };vector<node> G[maxn];int vis[maxn];int dis[maxn];void init() { for (int i = 0; i<maxn; i++) { G[i].clear(); dis[i] = INF; vis[i] = false; } } int Prim(int s) { priority_queue<node>Q; // 定义优先队列 int ans = 0; Q.push(node(s,0)); // 加入队列的起点 while (!Q.empty()) { node now = Q.top(); Q.pop(); // 取出距离最小的点 int v = now.v; if (vis[v]) continue; // 同一节点可能会推入两次或两次以上的队列，第一个标记后，剩下的需要直接跳过。 vis[v] = true; // 标记一下 ans += now.len; for (int i = 0; i<G[v].size(); i++) { // 开始更新 int v2 = G[v][i].v; int len = G[v][i].len; if (!vis[v2] && dis[v2] > len) { dis[v2] = len; Q.push(node(v2, dis[v2])); // 更新的点加入队列并排序 } } } return ans; }单源最短路
Dijkstra11.迪杰斯特拉算法
/* |Dijkstra算法| |适用于边权为正的有向图或无向图| |从单个源点到所有节点的最短路| |优化版:时间复杂度 O(elbn)| |16/11/05ztx， thanks to chaixiaojunstruct node { int v, len; node(int v = 0, int len = 0) :v(v), len(len) {} bool operator < (const node &a)const { // 从小到大的距离排序 return len > a.len; } }; vector<node>G[maxn]; bool vis[maxn]; int dis[maxn];void init() { for (int i = 0; i<maxn; i++) { G[i].clear(); vis[i] = false; dis[i] = INF; } } int dijkstra(int s, int e) { priority_queue<node>Q; Q.push(node(s, 0)); // 加入队列并进行排序 dis[s] = 0; while (!Q.empty()) { node now = Q.top(); // 取出目前最小的 Q.pop(); int v = now.v; if (vis[v]) continue; // 如果标记过， 直接continue vis[v] = true; for (int i = 0; i<G[v].size(); i++) { // 更新 int v2 = G[v][i].v; int len = G[v][i].len; if (!vis[v2] && dis[v2] > dis[v] + len) { dis[v2] = dis[v] + len; Q.push(node(v2, dis[v2])); } } } return dis[e]; }12.最短路径快速算法（Shortest Path Faster Algorithm）
/* |SPFA算法| |队列优化| 可处理负环|*/vector<node> G[maxn];bool inqueue[maxn];int dist[maxn];void Init() { for(int i = 0 ; i < maxn ; ++i){ G[i].clear(); dist[i] = INF; } } int SPFA(int s,int e) { int v1,v2,weight; queue<int> Q; memset(inqueue,false,sizeof(inqueue)); // 队列中是否有标记 memset(cnt,0,sizeof(cnt)); // 加入队列的次数 dist[s] = 0; Q.push(s); // 加入队列的起点 inqueue[s] = true; // 标记 while(!Q.empty()){ v1 = Q.front(); Q.pop(); inqueue[v1] = false; // 取消标记 for(int i = 0 ; i < G[v1].size() ; ++i){ // 搜索v1链表 v2 = G[v1][i].vex; weight = G[v1][i].weight; if(dist[v2] > dist[v1] + weight){ // 松弛操作 dist[v2] = dist[v1] + weight; if(inqueue[v2] == false){ // 再次加入队列 inqueue[v2] = true; //cnt[v2]++; // 判负环 //if(cnt[v2] > n) return -1; Q.push(v2); } } } } return dist[e]; }/* 不断地将s的邻接点加入队列，取出连续的松弛操作，直到队列为空 如果一个结点被加入队列超过n-1次，显然图中有负环 */13.弗洛伊德算法
/* |Floyd算法| |任意点对最短路算法| |图中任何两点的最短距离算法|*/for (int i = 0; i < n; i++) { // 初始化为0 for (int j = 0; j < n; j++) scanf("%lf", &dis[i][j]); } for (int k = 0; k < n; k++) { for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]); } }}14.染色法
/* |交叉染色法判断二分图| |16/11/05ztxint bipartite(int s) { int u, v; queue<int>Q; color[s] = 1; Q.push(s); while (!Q.empty()) { u = Q.front(); Q.pop(); for (int i = 0; i < G[u].size(); i++) { v = G[u][i]; if (color[v] == 0) { color[v] = -color[u]; Q.push(v); } else if (color[v] == color[u]) return 0; } } return 1; }15..匈牙利算法
/* |解决最大匹配问题| |实现递归| |16/11/05ztxvector<int>G[maxn]; bool inpath[maxn]; // 标记 int match[maxn]; // 记录匹配对象 void init() { memset(match, -1, sizeof(match)); for (int i = 0; i < maxn; ++i) { G[i].clear(); } } bool findpath(int k) { for (int i = 0; i < G[k].size(); ++i) { int v = G[k][i]; if (!inpath[v]) { inpath[v] = true; if (match[v] == -1 || findpath(match[v])) { // 递归 match[v] = k; // 也就是说，匹配对象是“k妹” return true; } } } return false; } void hungary() { int cnt = 0; for (int i = 1; i <= m; i++) { // m是需要匹配的“姐姐”数 memset(inpath, false, sizeof(inpath)); // 每次都要初始化 if (findpath(i)) cnt++; } cout << cnt << endl; }/* |解决最大匹配问题| |dfs实现| |16/11/05ztxint v1, v2; bool Map[501][501]; bool visit[501]; int link[501]; int result; bool dfs(int x) { for (int y = 1; y <= v2; ++y) { if (Map[x][y] && !visit[y]) { visit[y] = true; if (link[y] == 0 || dfs(link[y])) { link[y] = x; return true; } } } return false; } void Search() { for (int x = 1; x <= v1; x++) { memset(visit,false,sizeof(visit)); if (dfs(x)) result++; }}16.17.18背包问题
/* |01背包| 完全背包| |多重背包| |16/11/05ztx 01背包： void bag01(int cost,int weight) { for(i = v; i >= cost; --i) dp[i] = max(dp[i], dp[i-cost]+weight); } // 完全背包： void complete(int cost, int weight) { for(i = cost ; i <= v; ++i) dp[i] = max(dp[i], dp[i - cost] + weight); } // 多重背包： void multiply(int cost, int weight, int amount) { if(cost * amount >= v) complete(cost, weight); else{ k = 1; while (k < amount){ bag01(k * cost, k * weight); amount -= k; k += k; } bag01(cost * amount, weight * amount); } } // otherint dp[1000000];int c[55], m[110];int sum;void CompletePack(int c) { for (int v = c; v <= sum / 2; ++v){ dp[v] = max(dp[v], dp[v - c] + c); }}void ZeroOnePack(int c) { for (int v = sum / 2; v >= c; --v) { dp[v] = max(dp[v], dp[v - c] + c); }}void multiplePack(int c, int m） { if (m * c > sum / 2) CompletePack(c); else{ int k = 1; while (k < m){ ZeroOnePack(k * c); m -= k; k <<= 1; } if (m != 0){ ZeroOnePack(m * c); } }}19.最长上升子序列
/* |最长上升子序列| |状态转移| |16/11/05ztx* 状态转移dp[i] = max{ 1.dp[j] + 1 }; j<i; a[j]<a[i]; d[i]以i结尾最长的上升子序列 与i之前的 每个a[j]<a[i]的 j位置最长上升子序列+1后的值比*/void solve(){ // 参考挑战程序设计入门经典； for(int i = 0; i < n; ++i){ dp[i] = 1; for(int j = 0; j < i; ++j){ if(a[j] < a[i]){ dp[i] = max(dp[i], dp[j] + 1); } } }} /* 优化方法： dp[i]上升子序列的末尾元素表示长度为i+1 找第一个比dp末尾大的来代替 */ void solve() { for (int i = 0; i < n; ++i){ dp[i] = INF; } for (int i = 0; i < n; ++i) { *lower_bound(dp, dp + n, a[i]) = a[i]; // 返回一个指针 } printf("%d\n", *lower_bound(dp, dp + n, INF) - dp; }/* 函数lower_bound()回一个 iterator 它指向在[first,last)value可以插入标记的有序列，而不会破坏容器顺序的第一个位置，标记不小于value的值。*/20.最长的公共子序列
/* |最长的公共子序列| |递推形式| |16/11/05ztxvoid solve() { for (int i = 0; i < n; ++i) { for (int j = 0; j < m; ++j) { if (s1[i] == s2[j]) { dp[i + 1][j + 1] = dp[i][j] + 1; }else { dp[i + 1][j + 1] = max(dp[i][j + 1], dp[i + 1][j]); } } }}21.向量基本用法
/* |16/11/06ztxstruct node { double x; // 横坐标 double y; // 纵坐标 }; typedef node Vector;Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); } Vector operator - (Point A, Point B) { return Vector(A.x - B.y, A.y - B.y); } Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); } Vector operator / (Vector A, double p) { return Vector(A.x / p, A.y*p); } double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; } // 向量点乘 double Length(Vector A) { return sqrt(Dot(A, A)); } // 向量模长 double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); } // 夹角在向量之间 double Cross(Vector A, Vector B) { // 叉积计算 公式 return A.x*B.y - A.y*B.x; } Vector Rotate(Vector A, double rad) // 向量旋转 公式 { return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad)); } Point getLineIntersection(Point P, Vector v, Point Q, Vector w) { // T1是两条直线的交点 t2计算公式 Vector u = P - Q; double t = Cross(w, u) / Cross(v, w); // 是横坐标 return P + v*t; // 返回一个点 }22.寻求多边形面积
/* |16/11/06ztxnode G[maxn]; int n; double Cross(node a, node b) { // 叉积计算 return a.x*b.y - a.y*b.x; } int main() { while (scanf("%d", &n) != EOF && n) { for (int i = 0; i < n; i++) scanf("%lf %lf", &G[i].x, &G[i].y); double sum = 0; G[n].x = G[0].x; G[n].y = G[0].y; for (int i = 0; i < n; i++) { sum += Cross(G[i], G[i + 1]); } // 或者 //for (int i = 0; i < n; i++) { //sum += fun(G[i], G[（i + 1）% n]); //} sum = sum / 2.0; printf("%.1f\n", sum); } system("pause"); return 0; }23..判断线段相交
/* |16/11/06ztxnode P[35][105]; double Cross_Prouct(node A,node B,node C) { // 计算BA叉乘CA return (B.x-A.x)*(C.y-A.y)-(B.y-A.y)*(C.x-A.x); } bool Intersect(node A,node B,node C,node D) { // 判断线段是否交叉乘； if(min(A.x,B.x)<=max(C.x,D.x)&& // 快速排斥实验； min(C.x,D.x)<=max(A.x,B.x)&& min(A.y,B.y)<=max(C.y,D.y)&& min(C.y,D.y)<=max(A.y,B.y)&& Cross_Prouct(A,B,C)*Cross_Prouct(A,B,D)<0&& // 跨立实验； Cross_Prouct(C,D,A)*Cross_Prouct(C,D,B)<0) // 叉乘异号表示在两侧； return true; else return false; }24.求三角形外心
/* |16/11/06ztxPoint circumcenter(const Point &a, const Point &b, const Point &c) { ////返回三角形的外心 Point ret; double a1 = b.x - a.x, b1 = b.y - a.y, c1 = (a1*a1 + b1*b1) / 2; double a2 = c.x - a.x, b2 = c.y - a.y, c2 = (a2*a2 + b2*b2) / 2; double d = a1*b2 - a2*b1; ret.x = a.x + (c1*b2 - c2*b1) / d; ret.y = a.y + (a1*c2 - a2*c1) / d; return ret; }24.极角排序
/* |16/11/06ztxdouble cross(point p1, point p2, point q1, point q2) { // 叉积计算 return (q2.y - q1.y)*(p2.x - p1.x) - (q2.x - q1.x)*(p2.y - p1.y); } bool cmp(point a, point b) { point o; o.x = o.y = 0; return cross(o, b, o, a) < 0; // 叉积判断 } sort(convex + 1, convex + cnt, cmp); // 按角排序, 从小到大25.克努特-莫里斯-普拉特操作
/* |kmp算法| |字符串匹配| |17/1/21ztxvoid getnext(char str[maxn], int nextt[maxn]) { int j = 0, k = -1; nextt[0] = -1; while (j < m) { if (k == -1 || str[j] == str[k]) { j++; k++; nextt[j] = k; } else k = nextt[k]; }}void kmp(int a[maxn], int b[maxn]) { int nextt[maxm]; int i = 0, j = 0; getnext(b, nextt); while (i < n) { if (j == -1 || a[i] == b[j]) { // 母串不动，子串移动 j++; i++; } else { // i不需要追溯 // i = i - j + 1; j = nextt[j]; } if (j == m) { printf("%d\n", i - m + 1); // 母串的位置减去了子串的长度+1 return; } } printf("-1\n");}26.kmp扩展
/* |16/11/06ztx#include<iostream> #include<cstring> using namespace std;const int MM=100005; int next[MM],extand[MM]; char S[MM],T[MM]; void GetNext(const char *T) { int len = strlen(T),a = 0; next[0] = len; while(a < len - 1 && T[a] == T[a + 1]) a++; next[1] = a; a = 1; for(int k = 2; k < len; k ++) { int p = a + next[a] - 1,L = next[k - a]; if( (k - 1) + L >= p) { int j = (p - k + 1) > 0 ? (p - k + 1) : 0; while(k + j < len && T[k + j] == T[j]) j++; next[k] = j; a = k; }else next[k] = L; } } void GetExtand(const char *S,const char *T) { GetNext(T); int slen = strlen(S),tlen = strlen(T),a = 0; int MinLen = slen < tlen ? slen : tlen; while(a < MinLen && S[a] == T[a]) a++; extand[0] = a; a = 0; for(int k = 1; k < slen; k ++) { int p = a + extand[a] - 1, L = next[k - a]; if( (k - 1) + L >= p) { int j = (p - k + 1) > 0 ? (p - k + 1) : 0; while(k + j < slen && j < tlen && S[k + j] == T[j]) j ++; extand[k] = j; a = k; } else extand[k] = L; } } void show(const int *s,int len){ for(int i = 0; i < len; i ++) cout << s[i] << ' '; cout << endl; } int main() { while(cin >> S >> T) { GetExtand(S,T); show(next,strlen(T)); show(extand,strlen(S)); } return 0; }27.字典树
/* |16/11/06ztxstruct Trie{ int cnt; Trie *next[maxn]; Trie(){ cnt = 0; memset(next,0,sizeof(next)); } }; Trie *root; void Insert(char *word) { Trie *tem = root; while(*word != '\0') { int x = *word - 'a'; if(tem->next[x] == NULL) tem->next[x] = new Trie; tem = tem->next[x]; tem->cnt++; word++; } } int Search(char *word) { Trie *tem = root; for(int i=0;word[i]!='\0';i++) { int x = word[i]-'a'; if(tem->next[x] == NULL) return 0; tem = tem->next[x]; } return tem->cnt; } void Delete(char *word,int t) { Trie *tem = root; for(int i=0;word[i]!='\0';i++) { int x = word[i]-'a'; tem = tem->next[x]; (tem->cnt)-=t; } for(int i=0;i<maxn;i++) tem->next[i] = NULL; } int main() { int n; char str1[50]; char str2[50]; while(scanf("%d",&n)!=EOF) { root = new Trie; while(n--) { scanf("%s %sstr1，str2); if(str1[0]==i') { Insert(str2); }else if(str1[0] == 's') { if(Search(str2) printf("Yes\n"); else printf("No\n"); }else { int t = Search(str2); if(t) Delete(str2,t); } } } return 0; }28.AC自动机
/* |16/11/06ztx#include<iostream> #include<cstdio> #include<cstring> #include<string> using namespace std; #define N 1000010 char str[N], keyword[N]; int head, tail; struct node { node *fail; node *next[26]; int count; node() { //init fail = NULL;// 默认为空 count = 0; for(int i = 0; i < 26; ++i) next[i] = NULL; } }*q[N]; node *root; void insert(char *str) { // 建立Trie int temp, len; node *p = root; len = strlen(str); for(int i = 0; i < len; ++i) { temp = str[i] - 'a'; if(p->next[temp] == NULL) p->next[temp] = new node(); p = p->next[temp]; } p->count++; } void build_ac() { // fail指针初始化，BFS 数组模拟队列： q[tail++] = root; while(head != tail) { node *p = q[head++]; // 弹出队头 node *temp = NULL; for(int i = 0; i < 26; ++i) { if(p->next[i] != NULL) { if(p == root) { // 第一个元素fail必须指向根 p->next[i]->fail = root; }else { temp = p->fail; // 失败指针 while(temp != NULL) { // 两种情况结束：匹配为空或找到匹配 if(temp->next[i] != NULL) { // 找到匹配 p->next[i]->fail = temp->next[i]; break; } temp = temp->fail; } if(temp == NULL) // 为空而从头匹配 p->next[i]->fail = root; } q[tail++] = p->next[i]; // 入队 } } } } int query() // 扫描 { int index, len, result; node *p = root; // Tire入口 result = 0; len = strlen(str); for(int i = 0; i < len; ++i) { index = str[i] - 'a'; while(p->next[index] == NULL && p != root) // 指针跳转失败 p = p->fail; p = p->next[index]; if(p == NULL) p = root; node *temp = p; // p不动，后缀串计算temp while(temp != root && temp->count != -1) { result += temp->count; temp->count = -1; temp = temp->fail; } } return result; } int main() { int num; head= tail = 0; root = new node(); scanf("%d", &num); getchar(); for(int i = 0; i < num; ++i) { scanf("%s",keyword); insert(keyword); } build_ac(); scanf("%s", str); if(query()) printf("YES\n"); else printf("NO\n"); return 0; } /* 假设有N个模式串，平均长度为L；文章长度为M。 建立Trie树：O(N*L) 建立fail指针：O(N*L) 模式匹配：O(M*L) 因此，总时间复杂度为：O( (N+M)*L )。 建立Trie树：O(N*L) 建立fail指针：O(N*L) 模式匹配：O(M*L) 因此，总时间复杂度为：O( (N+M)*L )。*/29.线段树 1）点更新
/* |16/12/07ztxstruct node{ int left, right; int max, sum;};node tree[maxn << 2];int a[maxn];int n;int k = 1;int p, q;string str;void build(int m, int l, int r)//m 是 树的标号{ tree[m].left = l; tree[m].right = r; if (l == r){ tree[m].max = a[l]; tree[m].sum = a[l]; return; } int mid = (l + r) >> 1; build(m << 1, l, mid); build(m << 1 | 1, mid + 1, r); tree[m].max = max(tree[m << 1].max, tree[m << 1 | 1].max); tree[m].sum = tree[m << 1].sum + tree[m << 1 | 1].sum;}void update(int m, int a, int val)//a 是 节点位置， val 是 更新值(加减值){ if (tree[m].left == a && tree[m].right == a){ tree[m].max += val; tree[m].sum += val; return; } int mid = (tree[m].left + tree[m].right) >> 1; if (a <= mid){ update(m << 1, a, val); } else{ update(m << 1 | 1, a, val); } tree[m].max = max(tree[m << 1].max, tree[m << 1 | 1].max); tree[m].sum = tree[m << 1].sum + tree[m << 1 | 1].sum;}int querySum(int m, int l, int r){ if (l == tree[m].left && r == tree[m].right){ return tree[m].sum; } int mid = (tree[m].left + tree[m].right) >> 1; if (r <= mid){ return querySum(m << 1, l, r); } else if (l > mid){ return querySum(m << 1 | 1, l, r); } return querySum(m << 1, l, mid) + querySum(m << 1 | 1, mid + 1, r);}int queryMax(int m, int l, int r){ if (l == tree[m].left && r == tree[m].right){ return tree[m].max; } int mid = (tree[m].left + tree[m].right) >> 1; if (r <= mid){ return queryMax(m << 1, l, r); } else if (l > mid){ return queryMax(m << 1 | 1, l, r); } return max(queryMax(m << 1, l, mid), queryMax(m << 1 | 1, mid + 1, r));} build(1,1,n); update(1,a,b); query(1,a,b);2)区间更新
/* |16/11/06ztxtypedef long long ll; const int maxn = 100010; int t,n,q; ll anssum; struct node{ ll l,r; ll addv,sum; }tree[maxn<<2]; void maintain(int id) { if(tree[id].l >= tree[id].r) return ; tree[id].sum = tree[id<<1].sum + tree[id<<1|1].sum; } void pushdown(int id) { if(tree[id].l >= tree[id].r) return ; if(tree[id].addv){ int tmp = tree[id].addv; tree[id<<1].addv += tmp; tree[id<<1|1].addv += tmp; tree[id<<1].sum += (tree[id<<1].r - tree[id<<1].l + 1)*tmp; tree[id<<1|1].sum += (tree[id<<1|1].r - tree[id<<1|1].l + 1)*tmp; tree[id].addv = 0; } } void build(int id,ll l,ll r) { tree[id].l = l; tree[id].r = r; tree[id].addv = 0; tree[id].sum = 0; if(l==r) { tree[id].sum = 0; return ; } ll mid = (l+r)>>1; build(id<<1,l,mid); build(id<<1|1,mid+1,r); maintain(id); } void updateAdd(int id,ll l,ll r,ll val) { if(tree[id].l >= l && tree[id].r <= r) { tree[id].addv += val; tree[id].sum += (tree[id].r - tree[id].l+1)*val; return ; } pushdown(id); ll mid = (tree[id].l+tree[id].r)>>1; if(l <= mid) updateAdd(id<<1,l,r,val); if(mid < r) updateAdd(id<<1|1,l,r,val); maintain(id); } void query(int id,ll l,ll r) { if(tree[id].l >= l && tree[id].r <= r){ anssum += tree[id].sum; return ; } pushdown(id); ll mid = (tree[id].l + tree[id].r)>>1; if(l <= mid) query(id<<1,l,r); if(mid < r) query(id<<1|1,l,r); maintain(id); } int main() { scanf("%d",&t); int kase = 0 ; while(t--){ scanf("%d %d",&n,&q); build(1,1,n); int id; ll x,y; ll val; printf("Case %d:\n",++kase); while(q--){ scanf("%d",&id); if(id==0){ scanf("%lld %lld %lld",&x,&y,&val); updateAdd(1,x+1,y+1,val); } else{ scanf("%lld %lld",&x,&y); anssum = 0; query(1,x+1,y+1); printf("%lld\n",anssum); } } } return 0; }30.树状数组
/* |16/11/06ztx#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>using namespace std;typedef long long ll;const int maxn = 50005;int a[maxn];int n;int lowbit(const int t) { return t & (-t);}void insert(int t, int d) { while (t <= n){ a[t] += d; t = t + lowbit(t); }}ll getSum(int t) { ll sum = 0; while (t > 0){ sum += a[t]; t = t - lowbit(t); } return sum;}int main() { int t, k, d; scanf("%d", &t); k= 1; while (t--){ memset(a, 0, sizeof(a)); scanf("%d", &n); for (int i = 1; i <= n; ++i) { scanf("%d", &d); insert(i, d); } string str; printf("Case %d:\n", k++); while (cin >> str) { if (str == "End") break; int x, y; scanf("%d %d", &x, &y); if (str == "Query") printf("%lld\n", getSum(y) - getSum(x - 1)); else if (str == "Add") insert(x, y); else if (str == "Sub") insert(x, -y); } } return 0;}31.中国剩余定理(孙子定理)
/* |16/11/06ztxint CRT(int a[],int m[],int n) { int M = 1; int ans = 0; for(int i=1; i<=n; i++) M *= m[i]; for(int i=1; i<=n; i++) { int x, y; int Mi = M / m[i]; extend_Euclid(Mi, m[i], x, y); ans = (ans + Mi * x * a[i]) % M; } if(ans < 0) ans += M; return ans; } void extend_Euclid(int a, int b, int &x, int &y) { if(b == 0) { x = 1; y = 0; return; } extend_Euclid(b, a % b, x, y); int tmp = x; x = y; y = tmp - (a / b) * y; }
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