Protecting the Flowers

这道题目的难点就是如何设置这些牛的优先级，如果单纯只看一头牛破坏的花的数量，那么有一头牛花费的时间极大（无穷？）而破坏花的数目仅仅只是比其他牛破坏的数目略大，则这样的优先级是不恰当的，只看时间，也能举出这样的反例，那就只能把时间和破坏的数目综合起来看，举个例子，只比较两头牛1和2，t*（d1+d2）+t2*d1(先牵走牛2）、t*（d1+d2）+t1*d2（先牵走牛1） t*（d1+d2）为牵走其它牛时这两头牛破坏的花的数目，比较这两个数字，较大的即破坏花的数目较多的应该优先牵走，假设t*（d1+d2）+t2*d1>t*（d1+d2）+t1*d2 则化简为 t2*d1>t1*d2 移项 d1/t1>d2/t2,也就是d和t这两个比值大的优先级高

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 T_i minutes (1 ≤ T_i ≤ 2,000,000) away from its own barn. Furthermore, while waiting for transport, she destroys D_i (1 ≤ D_i ≤ 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 × T_i minutes (T_i to get there and T_i 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, T_iand D_i, that describe a single cow's characteristics

Output

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

Sample Input

63 12 52 33 24 11 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.

#include<stdio.h>#include<queue>using namespace std;struct list{double t;double d;double dive;friend bool operator <(struct list a,struct list b){return a.dive<b.dive;}}; int main(){int n;scanf("%d",&n);struct list m;priority_queue<struct list> q;long long sum,res;sum=0,res=0;for(int i=0;i<n;i++){scanf("%lf%lf",&m.t,&m.d);sum+=m.d;m.dive=m.d/m.t;q.push(m); }while(!q.empty()){sum-=q.top().d;res+=2*q.top().t*sum;q.pop();}printf("%lld",res);return 0;}