POJ 3262 Protecting the Flowers 贪心比率

本文主要是介绍POJ 3262 Protecting the Flowers 贪心比率，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

Description
Farmer John went to cut some wood and left N (2 ≤ N ≤ 100,000) cows eating the grass, as usual. When he returned, he found to his horror that the cluster of cows was in his garden eating his beautiful flowers. Wanting to minimize the subsequent damage, FJ decided to take immediate action and transport each cow back to its own barn.
Each cow i is at a location that is Ti minutes (1 ≤ Ti ≤ 2,000,000) away from its own barn. Furthermore, while waiting for transport, she destroys Di (1 ≤ Di ≤ 100) flowers per minute. No matter how hard he tries, FJ can only transport one cow at a time back to her barn. Moving cow i to its barn requires 2 × Ti minutes (Ti to get there and Ti to return). FJ starts at the flower patch, transports the cow to its barn, and then walks back to the flowers, taking no extra time to get to the next cow that needs transport.
Write a program to determine the order in which FJ should pick up the cows so that the total number of flowers destroyed is minimized.
Input
Line 1: A single integer N
Lines 2…N+1: Each line contains two space-separated integers, Ti and Di, that describe a single cow’s characteristics
Output
Line 1: A single integer that is the minimum number of destroyed flowers
Sample Input
6
3 1
2 5
2 3
3 2
4 1
1 6
Sample Output
86
Hint
FJ returns the cows in the following order: 6, 2, 3, 4, 1, 5. While he is transporting cow 6 to the barn, the others destroy 24 flowers; next he will take cow 2, losing 28 more of his beautiful flora. For the cows 3, 4, 1 he loses 16, 12, and 6 flowers respectively. When he picks cow 5 there are no more cows damaging the flowers, so the loss for that cow is zero. The total flowers lost this way is 24 + 28 + 16 + 12 + 6 = 86.

题目大意：
花园中有n头牛，将牛 i 赶回相应的畜栏需要用时 ti，如果不赶该牛，则每单位时间它吃掉 di 朵花，每次只能赶一头牛，注意将牛赶回后还需要回到花园，因此总用时2*ti，求将所有牛赶回之后，花朵的最小消耗。

解题思路：
每次选择当前的最优解，贪心思想。
在这里插入图片描述

AC代码

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
struct node {int t, cost;
}list[100001];
bool cmp(node a, node b) {return 1.0* a.t / a.cost < 1.0* b.t / b.cost;//注意计算精度！！！
}
int main() {int n;while (cin >> n) {long long ans = 0; long long all = 0;//注意数据范围for (int i = 1; i <= n; i++) {scanf("%d%d", &list[i].t, &list[i].cost);all += list[i].cost;}sort(list + 1, list + 1 + n, cmp);for (int i = 1; i <= n; i++) {ans += (all - list[i].cost)*list[i].t * 2;all -= list[i].cost;}cout << ans << endl;}
}