题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1114
题目大意:给定一个储蓄罐可容纳的重量和n个价值为p重量为w的硬币,问在填满储蓄罐的情况最小的价值为多少?如果没办法填满输出This is impossible.
解题思路:因为题目的数据范围很小,时间又给很多,可以直接暴力,枚举每次选的数量后相当于01背包。状态转移方程:dp[j] = min(dp[j-k*w[i]] + k * p[i]) (dp[j]为容量为j时的最小价值,k 《 num[i],p[i]为价值,w[i]为重量)
测试数据:
3
10 110
2
1 1
30 50
10 110
2
1 1
50 30
1 6
2
10 3
20 4
代码:
#include <stdio.h> #include <string.h> #define INF -2139062144 int main() { int t,i,j,n,k,sum,pig; int w[600],v[600],dp[10001]; scanf("%d",&t); while (t--){ scanf("%d%d",&pig,&sum); memset(dp,128,sizeof(dp)); dp[0] = 0,sum -= pig; scanf("%d",&n); for (i = 1; i <= n; ++i) scanf("%d%d",&w[i],&v[i]); for (i = 1; i <= n; ++i) for (j = v[i]; j <= sum; j ++) if (dp[j - v[i]] != INF){ if (dp[j] != INF) dp[j] = dp[j] < dp[j-v[i]] + w[i] ? dp[j] : dp[j-v[i]] + w[i]; else dp[j] = dp[j-v[i]] + w[i]; } if (dp[sum] == INF || sum == 0) printf("This is impossible.\n"); else printf("The minimum amount of money in the piggy-bank is %d.\n",dp[sum]); } }
本文ZeroClock原创,但可以转载,因为我们是兄弟。