POJ 1384 Piggy-Bank 背包DP

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所谓的完全背包，就是说物品没有限制数量的。

怎么起个这么intimidating（吓人）的名字？

其实和一般01背包没多少区别，不过数量可以无穷大，那么就可以利用一个物品累加到总容量结尾就可以了。

本题要求装满的，故此增加个限制就可以了。

#include <stdio.h>#include <stdlib.h>#include <string.h>inline int min(int a, int b) { return a < b? a : b; }const int MAX_W = 10001;const int MAX_N = 501;int val[MAX_N] = {0};int wei[MAX_N] = {0};int tbl[MAX_W];int bagDP(int N, int W){for (int i = 0; i <= W; i++) tbl[i] = 0;for (int j = wei[1]; j <= W; j += wei[1])tbl[j] = val[1] + tbl[j-wei[1]];for (int i = 2; i <= N; i++){for (int j = wei[i]; j <= W; j++){if (j-wei[i]==0 || tbl[j-wei[i]]){if (tbl[j]) tbl[j] = min(tbl[j],tbl[j-wei[i]]+val[i]);else tbl[j] = tbl[j-wei[i]]+val[i];}}}return tbl[W];}int main(){int E, F;//weight of an empty pig and of the pig filled with coinsint T, N;// P:value, W: weightscanf("%d", &T);while (T--){scanf("%d %d %d", &E, &F, &N);for (int i = 1; i <= N; i++){scanf("%d %d", &val[i], &wei[i]);}int ans = bagDP(N, F-E);if (ans) printf("The minimum amount of money in the piggy-bank is %d.\n", ans);else puts("This is impossible.");}return 0;}

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