UVA - 757 Gone Fishing 贪心+枚举

来源：互联网发布：单片机液晶屏与电机编辑：程序博客网时间：2024/05/17 22:15

题目大意：有n个湖泊，每个湖泊最初的5分钟能钓到f条鱼，每五分钟减少d条鱼，鱼的数目不能小于d也不能为负数，求在h小时能钓到的鱼的最大数目和在每个池塘带了多少分钟

解题思路：一个个枚举，如果用总时间减去到达另一个湖泊的时间的话，就表示它可以在两个湖泊随意行走了，然后在这些时间找到优解，并和前几次的最优解进行比较，就能得到结果

#include<cstdio>#include<cstring>const int maxn = 30 + 5;int n, h;int r[maxn];struct Lake{int f;int d;int t;};Lake l1[maxn],l2[maxn],out[maxn];int main(){int b = 1;while(scanf("%d", &n) && n) {if(b) b = 0;elseprintf("\n");scanf("%d", &h);h = h * 60;for(int i = 0; i < n; i++)  {scanf("%d", &l1[i].f);l1[i].t = 0;}for(int i = 0; i  < n; i++)scanf("%d", &l1[i].d);for(int i = 0; i < n - 1; i++)scanf("%d", &r[i]);int max = -0x3f3f3f3f;int count = 0;for(int i = 0 ; i < n; i++) {int sum = 0;int time = h;for(int j = 0; j < n; j++)  {l2[j] = l1[j];}for(int j = 0; j < i; j++)time = time - 5 * r[j];while(time > 0) {int m = -0x3f3f3f3f;int s;for(int j = 0; j <= i; j++)if(l2[j].f > m ) {m = l2[j].f;s = j;}sum = sum + l2[s].f;if(l2[s].f - l2[s].d >= 0)l2[s].f = l2[s].f - l2[s].d;elsel2[s].f = 0;l2[s].t = l2[s].t + 5;time = time -5;}if(sum > max) {max = sum;for(int j = 0; j < n; j++)  out[j] = l2[j];}}printf("%d",out[0].t);for(int i = 1 ; i < n; i++)printf(", %d",out[i].t);printf("\n");printf("Number of fish expected: %d\n",max);}return 0;}

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