ZOJ-2972

来源：互联网发布：sql server数据库知识编辑：程序博客网时间：2024/06/01 14:15

继续刷DP，这题没有思路，看了看别人的题解，才知道可以根据每轮所剩体力值来DP。。知道这点之后就简单了，dp[i][j]数组的值表示第i个栏时，剩j点体力所需的最小时间，然后再循环推最优解

#include<stdio.h>#include<string.h>unsigned min(unsigned a, unsigned b){return a < b ? a : b;}int main(){int t, n, m;unsigned dp[111][111];scanf("%d", &t);while (t--){scanf("%d %d", &n, &m);unsigned i, j, t1, t2, t3, f1, f2, temp, res;memset(dp, -1, sizeof(dp));memset(dp[0], 0, sizeof(dp[0]));for (i = 1; i <= n; i++){scanf("%u %u %u %u %u", &t1, &t2, &t3, &f1, &f2);for (j = 0; j <= m; j++){dp[i][j] = min(dp[i - 1][j] + t2, dp[i][j]);if (j >= f1)dp[i][j - f1] = min(dp[i - 1][j] + t1, dp[i][j - f1]);temp = min(j + f2, m);dp[i][temp] = min(dp[i - 1][j] + t3, dp[i][temp]);}}res = 0xFFFFFFFF;for (j = 0; j <= m; j++)res = min(res, dp[n][j]);printf("%u\n", res);}return 0;}

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