POJ3262贪心

来源:互联网 发布:java http协议面试题 编辑:程序博客网 时间:2024/06/07 07:14
Protecting the Flowers
Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 7263 Accepted: 2938

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 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.

Input

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

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


    题意:有一群牛在吃草,现在需要把牛全部运走,每次运走一头牛。现给出运每头牛需要的时间与每头牛每分钟吃草数。求最小的吃草值。
    解析:首先得明白,既然要求所吃草最少,就得满足——优先把吃草速度快的牛运走。那么贪心策略救出来了。按照吃草的速度给所有的数据排序,然后再依次进行运算。以下是我的代码
#include <stdio.h>#include <algorithm>#include <iostream>using namespace std;typedef struct cow{    int minute;    int weight;} T;T arr[100005];int cmp(T A,T B){    return (1.0*B.weight/B.minute)<(1.0*A.weight/A.minute);}int main(){    //freopen("in.txt","r",stdin);    int n;    scanf("%d",&n);    long long s=0,sum=0;    for(int i=0; i<n; i++)    {        scanf("%d%d",&arr[i].minute,&arr[i].weight);        s+=arr[i].weight;    }    sort(arr,arr+n,cmp);    for(int i=0; i<n; i++)    {        s-=arr[i].weight;        sum+=s*2*arr[i].minute;    }    printf("%lld\n",sum);    return 0;}

原创粉丝点击