题目链接：https://ac.nowcoder.com/acm/contest/984/I/
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld

题目描述

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.

输入描述

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

输出描述

Line 1: A single integer that is the minimum number of destroyed flowers

输入

6
3 1
2 5
2 3
3 2
4 1
1 6

输出

86

解题思路

题意:有n头牛在吃花，每头牛牵走（避免吃花）要花费2*t秒（来回），每头牛吃花的速率是d，要你自己设计一个牵走牛的顺序，使被牛吃的花最少?
思路:贪心思想，设a，b是两头牛，我们想要的就是先送损失的花少的。
送a牛则b牛损失的花为2*t_a*d_b，送b牛则a牛损失的花为 2*t_b*d_a；
假设b牛吃得花比a牛多，那么就有关系式t_a*d_b < t_b*d_a，即d_b/t_b > d_a/t_a，所以这样就可以按照这个速率比时间的值来贪心，让被牛吃的花尽量少。

Accepted Code:

//正着算 
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
struct edge {
    int t, c;
    bool operator < (const edge & r) const {
        return t * r.c < r.t * c;
    }
}a[N];
int main() {
    int n;
    scanf("%d", &n);
    long long ans = 0, sum = 0;
    for(int i = 0; i < n; i++) {
    	scanf("%d%d", &a[i].t, &a[i].c);
    	sum += a[i].c;
    }
    sort(a, a + n);
    for (int i = 0; i < n; i++) {
    	sum -= a[i].c;
        ans += 2ll * a[i].t * sum;
    }
    printf("%lld\n", ans);
    return 0;
}
//反着算
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
struct edge {
    int t, c;
    bool operator < (const edge & r) const {
        return t * r.c > r.t * c;
    }
}a[N];
int main() {
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
        scanf("%d%d", &a[i].t, &a[i].c);
    sort(a, a + n);
    long long ans = 0, sum = 0;
    for (int i = 0; i < n; i++) {
        ans += 2ll * a[i].t * sum;
        sum += a[i].c;
    }
    printf("%lld\n", ans);
    return 0;
}