POJ

题目链接：http://poj.org/problem?id=3262点击打开链接

Protecting the Flowers

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 7870 Accepted: 3175

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

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.

Source

USACO 2007 January Silver

给你一群牛的赶回家的时间和每分钟踩花的数量 让你求最少的踩花数

从整体来看可以知道 将所有牛赶回去的时间固定 因此我们需要将每分钟踩花最多 也就是破坏力最强的牛先赶回家 

因此计算出每头牛的破坏力然后从大到小排序赶回家即可

其他博客有分d/t与t/d 不太清楚 觉得这样思考就很合理

注意两个计算方式的排序顺序不同

#include <iostream>#include <stdio.h>#include <limits.h>#include <stack>#include <algorithm>#include <queue>#include <string.h>#include <set>using namespace std;struct xjy{    double k;    int  t;    int d;    bool operator < (const xjy &r)const    {        return k<r.k;    }};vector <xjy > s;int main(){    int n;    long long int sum=0;    long long int ans=0;    scanf("%d",&n);    for(int i=1;i<=n;i++)    {        xjy mid;        scanf("%d%d",&mid.t,&mid.d);        mid.k=mid.d*1.0/mid.t;        s.push_back(mid);        sum+=mid.d;    }    sort(s.begin(),s.end());    reverse(s.begin(),s.end());    for(int i=0;i<s.size();i++)    {        sum-=s[i].d;        ans+=(2*s[i].t*sum);    }    cout << ans;}