pku 1928 The Peanuts

题目意思：

求dodo在不大于K的时间内能采到的peanuts最大数量，其中(1)走一格，(2)从马路进入田地，(3)从田地进入马路，(4)采一处peanuts。都消耗单位时间1。dodo只能从大到小采。

 

解题思路：

纯模拟，注意细节就行。

 

代码：

#include <stdio.h>#include <stdlib.h>#include <math.h>typedef struct{int x,y;int amount;}Peanut;int cmp(const void *a, const void *b){return ((Peanut *)b)->amount - ((Peanut *)a)->amount;}int distance(Peanut a, Peanut b){return abs(a.x - b.x) + abs(a.y - b.y);}Peanut peanuts[2500];int M,N,K,P;int main(){int i,j,num,t,sum,pick,temp;scanf("%d", &t);while(t--){scanf("%d%d%d", &M, &N, &K);P = 0;for(i = 1; i <= M; ++i){for(j = 1; j <= N; ++j){scanf("%d", &num);if(num){peanuts[P].amount = num;peanuts[P].x = i;peanuts[P].y = j;++P;}}}qsort(peanuts, P, sizeof(Peanut), cmp);sum = 0;pick = 0;temp = peanuts[0].x + 1 + peanuts[0].x; // 去 + 采 + 回if(temp <= K){pick += peanuts[0].amount;sum = temp - peanuts[0].x; // 第一个特殊处理 for(i = 1; i < P; ++i){// temp = 已走 + 去 + 采 + 回 temp = sum + distance(peanuts[i], peanuts[i - 1]) + 1 + peanuts[i].x;if (temp <= K){ // 如果这个过程时间允许，就去采 pick += peanuts[i].amount;sum = temp - peanuts[i].x;}else{ // 否则退出 break;}}} printf("%d/n", pick);}return 0;}