完全背包问题，附上例题(HDU

完全背包问题的题目样式

　　有 N 种物品和一个容量为 W 的背包，每种物品都有无限件可用。放入第 i 种物品 的费用是 Vi，价值是 Wi。求解：将哪些物品装入背包，可使这些物品的耗费的费用总和不超过背包容量，且价值总和最大。

　　这个问题类似于 0-1 背包问题，所不同的是每种物品有无限件。也就是从每种物品的角度考虑，与它相关的策略已并非取或不取两种，而是有取 0 件、取 1 件、取 2 件等。所以我们这里进行的策略是在之前选取过此物品后继续选取此物品进行尝试 

for（i=0  -  N）

　　for(j=w[i]  -  M)

　　　　dp[i][j] = max(dp[i-1][j],dp[i-1][j-w[i]]+v[i]);

这题是求最小值

Piggy-Bank　　　　http://acm.hdu.edu.cn/showproblem.php?pid=1114

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27240    Accepted Submission(s): 13780

Problem Description

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! 

 

Input

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. 

 

Output

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.". 

 

Sample Input

3

10 110

2

1 1

30 50

10 110

2

1 1

50 30

1 6

2

10 3

20 4

 

Sample Output

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.

 

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;
const int INF = 0x3f3f3f3f;
int main()
{
    int T;
    while(scanf("%d",&T) != EOF) {
        while(T--) {
            int w1, w2, Space;
            scanf("%d%d",&w1,&w2);
            Space = w2 - w1;
            int n;
            int dp[11000];
            dp[0] = 0;
            for(int i = 1 ; i <= Space ;i++)
                dp[i] = INF;
            scanf("%d",&n);
            int v[600], w[600];
            for(int i = 0; i < n; i++)
                scanf("%d%d",&v[i],&w[i]);
            for(int i = 0 ; i < n; i++) {
                for(int j = w[i]; j <= Space; j++)
                    dp[j] = min(dp[j],dp[j-w[i]]+v[i]);
            }
            if(dp[Space] == 0x3f3f3f3f) printf("This is impossible.
");
            else printf("The minimum amount of money in the piggy-bank is %d.
",dp[Space]);
        }

    }
    return 0;
}