POJ 3262 Protecting the Flowers 贪心（性价比）

Protecting the Flowers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7812		Accepted: 3151

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 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_i and 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

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.

Source

USACO 2007 January Silver

 

 

题意：农夫John养的植物被cow吃了（John和cow的日常。。）现在给我们n组数，每组数分别表示赶走cow的单程时间和该cow每单位时间吃掉的植物数，求怎样赶cow使得损失最少。

思路：贪心。先赶走的cow一定是当前牛群里，用时尽可能少，吃得尽可能多的一头，但不一定是最少和最多。所以可以把t和d以比值的方式记录下来（性价比），这样就可以同时把t、d作为参数来考虑。每次先赶当前t/d最小的，使得总损失最小。

 

 

#include<stdio.h>
#include<algorithm>
using namespace std;

struct Node{
    double t,d;
}node[];
bool cmp(Node a,Node b)
{
    return (a.t/a.d)<(b.t/b.d);  //贪心
}
int main()
{
    int n,i;
    long long sum,ans;
    scanf("%d",&n);
    sum=;
    for(i=;i<=n;i++){
        scanf("%lf%lf",&node[i].t,&node[i].d);
        sum+=node[i].d;
    }
    sort(node+,node+n+,cmp);
    ans=;
    for(i=;i<=n;i++){
        sum-=node[i].d;
        ans+=sum*node[i].t*;
    }
    printf("%lld\n",ans);
    return ;
}