POJ 1384 Piggy-Bank（DP完全背包）

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#pragma warning(disable:4996)#include <cstdio>#include <cstring>#include <algorithm>#define LL long longusing namespace std;//很简单的完全背包问题，MLE了//复习了一下空间的优化 ^_^ ~ ~LL dp[10005];int w[505], p[505];int main(){int t; scanf("%d", &t);while (t--){memset(dp, 0, sizeof dp);int E, F; scanf("%d %d", &E, &F);int tol = F - E;int n; scanf("%d", &n);for (int i = 1; i <= n; i++)scanf("%d %d", p + i, w + i);/*for (int j = 1; j <= n; j++)dp[0][j] = 0;for (int i = 1; i <= tol; i++)dp[i][1] = (i >= w[1] && i%w[1] == 0) ? (i / w[1] * p[1]) : (LL(50000)*10000+1);for (int j = 2; j <= n; j++){for (int i = 1; i < w[j]; i++)dp[i][j] = dp[i][j - 1];for (int i = w[j]; i <= tol; i++){dp[i][j] = min(dp[i - w[j]][j] + p[j], dp[i][j - 1]);}}*///dp[i]表示 i 的重量最少能有多少钱//初始化时是没有钱币时，所以都是impossiblefor (int i = 1; i <= tol; i++)dp[i] = (LL)50000 * 10000 + 1;for (int j = 1; j <= n; j++){//这句话可以省略，因为本层的dp[i]跟上一层的dp[i]没有变化//for (int i = 1; i < w[j]; i++)dp[i] = dp[i];for (int i = w[j]; i <= tol; i++) {dp[i] = min(dp[i - w[j]] + p[j], dp[i]);}}if (dp[tol] == (LL)50000 * 10000 + 1){printf("This is impossible.\n");}else{printf("The minimum amount of money in the piggy-bank is %I64d.\n", dp[tol]);}}return 0;}

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