HDU1114 Piggy-Bank完全背包变形

Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small money, he takes all the coins and throws them into a piggy-bank. You know that this process is irreversible, the coins cannot be removed without breaking the pig. After a sufficiently long time, there should be enough cash in the piggy-bank to pay everything that needs to be paid.

But there is a big problem with piggy-banks. It is not possible to determine how much money is inside. So we might break the pig into pieces only to find out that there is not enough money. Clearly, we want to avoid this unpleasant situation. The only possibility is to weigh the piggy-bank and try to guess how many coins are inside. Assume that we are able to determine the weight of the pig exactly and that we know the weights of all coins of a given currency. Then there is some minimum amount of money in the piggy-bank that we can guarantee. Your task is to find out this worst case and determine the minimum amount of cash inside the piggy-bank. We need your help. No more prematurely broken pigs!

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing two integers E and F. They indicate the weight of an empty pig and of the pig filled with coins. Both weights are given in grams. No pig will weigh more than 10 kg, that means 1 <= E <= F <= 10000. On the second line of each test case, there is an integer number N (1 <= N <= 500) that gives the number of various coins used in the given currency. Following this are exactly N lines, each specifying one coin type. These lines contain two integers each, Pand W (1 <= P <= 50000, 1 <= W <=10000). P is the value of the coin in monetary units, W is it’s weight in grams.

Print exactly one line of output for each test case. The line must contain the sentence “The minimum amount of money in the piggy-bank is X.” where X is the minimum amount of money that can be achieved using coins with the given total weight. If the weight cannot be reached exactly, print a line “This is impossible.”.

3 10 110 2 1 1 30 50 10 110 2 1 1 50 30 1 6 2 10 3 20 4

The minimum amount of money in the piggy-bank is 60. The minimum amount of money in the piggy-bank is 100. This is impossible.

背包完全变形，在装满背包时寻求最小值，并引用背包九个字

在我们看到的背包问题中，实际上有两种不同的问题。有的题目要求“正好装满背包”时的最佳解决方案，有的题目不要求背包装满。区分这两种问题的实现方法在初始化时是不同的。 如果是第一个要求背包装满的问题，那么在初始化中，除了F[0]0，其他F[1..V]均设为−∞，这样就可以保证最终获得的F[V]正好装满背包的最佳解决方案。 如果不要求背包装满，只希望价格尽可能大，F[0应该初始化..V]全部设为0。 为什么会这样？可以理解的是，初始化的F数组实际上是在没有任何物品可以放入背包的情况下的合法状态。如果要求背包装满，此时只有容量为0的背包可以“装满”而不装满任何东西，价值为0。其他容量的背包没有合法的解决方案，属于未定义的状态，应该赋予价值-∞了。如果背包不必装满，那么任何容量的背包都有一个合法的解决方案，“什么都不装”，这个解决方案的价值是0，所以初始状态的价值都是0。 这个小技巧可以完全推广到其他类型的背包问题上，状态转移前的初始化就不解释了。

因为这个问题需要的是满时的最小值，所以初始化时给予的是无限的，要求min

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>using namespace std;int dp[10005];int w[10005];int val[10005];const int maxn = 9999999;int main(){   int t;   scanf("%d",&t);   while (t--)   {        for (int i = 0;i<10005;i++)          dp[i] = maxn;        int e,f;        scanf("%d%d",&e,&f);        int v = f - e;        int n;        scanf("%d",&n);        for (int i = 0;i<n;i++)          scanf("%d%d",&val[i],&w[i]);        dp[0] = 0;        for (int i = 0;i<n;i++)           for (int j = w[i];j<=v;j++)             dp[j] = min(dp[j],dp[j-w[i]]+val[i]);        if (dp[v] == 9999999)        printf("This is impossible.\n");        else        printf("The minimum amount of money in the piggy-bank is %d.\n",dp[v]);        memset(val,0,sizeof(val));        memset(w,0,sizeof(w));   }}