POJ_3262_Protecting the Flowers(greedy)

Protecting the Flowers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4950		Accepted: 1975

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 destroysD_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 cowi to its barn requires 2 × T_i minutes (T_i to get there andT_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.

题意：农夫养的N头牛在他的花园里吃花！这让农夫很苦恼，他一次只能把一头牛赶回牛栏，现在给你赶回每头牛到牛栏的时间Ti(也就是说来回要2*Ti的时间)，以及这只牛每分钟能吃掉花的数量Di，求这些牛最少能吃掉多少花。

分析：贪心题。我们拿出两头牛a，b来考虑。若先把牛a赶走，那么吃掉的花的数量是2*(Ta)*(Db)；若先把牛b赶回牛栏，那么吃掉花的数量是2*(Tb)*(Da)。所以最优化原则就是取min(2*(Ta)*(Db),2*(Tb)*(Da))。以此类推，贪心规则就出来了，按照此方法sort一遍就行了。

题目链接： http://poj.org/problem?id=3262

代码清单：

#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<cctype>
#include<string>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;

const int maxn = 100000 + 5;
const int maxv = 200000 + 5;

struct Point{
    int T;
    int D;
}point[maxn];

int N;

bool cmp(Point a,Point b){
    return a.D*b.T>a.T*b.D;
}

void input(){
    scanf("%d",&N);
    for(int i=0;i<N;i++){
        scanf("%d%d",&point[i].T,&point[i].D);
    }
}

void solve(){
    sort(point,point+N,cmp);
    ll ans=0,t=0;
    for(int i=0;i<N;i++){
        ans=ans+(ll)point[i].D*t;
        t=t+(ll)(2*point[i].T);
    }
    printf("%I64d\n",ans);
}

int main(){
    input();
    solve();
    return 0;
}