HDU 1114 Piggy-Bank（完全背包）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1114

题目大意：根据储钱罐的重量，求出里面钱最少有多少。给定储钱罐的初始重量，装硬币后重量，和每个对应面值硬币的重量。

Sample Input

10 110

1 1

30 50

10 110

1 1

50 30

1 6

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.

分析：这道题目的动态规划思想很简单，就是背包。

代码如下：

 1 # include<stdio.h>
 2 # include<string.h>
 3 const int N = 501;
 4 const int INF = 0xffffff;
 5 int f[10001];    //f[i]表示总重量为i的硬币，价值为多少
 6 int p[N],w[N];
 7 int main()
 8 {
 9     int T;
10     scanf("%d",&T);
11     while (T--){
12         int n,a,b,i,j;
13         scanf("%d%d",&a,&b);
14         f[0] = 0;    //表示重量为0的硬币价值为0
15         for (i=1;i<=b;i++) f[i]=INF;
16         scanf("%d",&n);
17         for (i=1;i<=n;i++) scanf("%d%d",&p[i],&w[i]);
18         for (i=1;i<=n;i++){
19             for (j=w[i];j<=b-a;j++){
20                 if (f[j-w[i]]+p[i]<f[j]) f[j]=f[j-w[i]]+p[i];
21             }
22         }
23         if (f[b-a]==INF) printf("This is impossible.
");
24         else printf("The minimum amount of money in the piggy-bank is %d.
",f[b-a]);
25     }
26     return 0;
27 }