POJ 1384 Piggy-Bank 背包DP

所谓的全然背包，就是说物品没有限制数量的。

怎么起个这么intimidating（吓人）的名字？

事实上和一般01背包没多少差别，只是数量能够无穷大，那么就能够利用一个物品累加到总容量结尾就能够了。

本题要求装满的，故此添加个限制就能够了。

#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 coins
	int T, N;// P:value, W: weight
	scanf("%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;
}