链接:https://ac.nowcoder.com/acm/problem/25043

来源:牛客网

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
示例1
输入
6
3 1
2 5
2 3
3 2
4 1
1 6
输出
86

Hint
Because the number of flowers that are destroyed by a couple of cows are divided from others.
So we just need to compare the neighboring two one by one and sort them.
Then we calculate the sum of destroyed flowers.
It will be the answer.

#include <bits/stdc++.h>

using namespace std;

struct node
{
    int t,d;
}cow[100000];

bool cmp(node a,node b)
{
    return a.t*b.d<a.d*b.t;//compare
}

int main()
{
    int N;
    scanf("%d",&N);
    for(int i=0;i<N;i++)//input
    {
        scanf("%d%d",&cow[i].t,&cow[i].d);
    }
    sort(cow,cow+N,cmp);
    long long int sum=0,time=0;
    for(int i=0;i<N;i++)
    {
        sum +=time*cow[i].d;
        time +=cow[i].t*2;
    }
    printf("%lld",sum);
    return 0;
}