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poj-1384 Piggy-Bank 地址:http://poj.org/problem?id=1384
题意:
知道盒子里面的物体的总重量,得到每一种硬币的价格和重量,求最少钱构成盒子物体总重量的钱的数量。
分析:
典型的不限重背包问题。当然维度也是在可以接受的范围, 直接设置dp数组进行min操作。
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <string> using namespace std; const int maxn = 505; const int max_val = 10000; int E,F, n, dp[max_val+5]; struct Node{ int p,w; }nd[maxn]; int cmp(const void* a, const void* b){ Node* aa = (Node *)a; Node* bb = (Node *)b; return (bb->w*1.0)/(bb->p*1.0) - (aa->w*1.0)/(aa->p*1.0); } int main(){ freopen("in.txt", "r", stdin); int test_num, i, j; scanf("%d", &test_num); while(test_num--){ scanf("%d %d", &E, &F); scanf("%d", &n); for(i=0; i<n; i++){ scanf("%d %d", &nd[i].p, &nd[i].w); } memset(dp, 0x3f3f3f3f, sizeof(dp)); dp[0] = 0; for(i=0; i<n; i++){ for(j=nd[i].w; j<=(F-E); j++){ if(dp[j - nd[i].w] != 0x3f3f3f3f ){ dp[j] = min(dp[j-nd[i].w]+nd[i].p, dp[j]); } } } if(dp[F-E] == 0x3f3f3f3f){ printf("This is impossible.\n"); }else{ printf("The minimum amount of money in the piggy-bank is %d.\n", dp[F-E]); } } return 0; }
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原文地址:http://www.cnblogs.com/zhang-yd/p/5811805.html