[贪心] poj3262 Protecting the Flowers

Protecting the Flowers
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 5343 Accepted: 2105
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

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.

有n个牛在FJ的花园乱吃。
所以FJ要赶他们回牛棚。
每个牛在被赶走之前每秒吃Di个花朵。赶它回去FJ要花的时间是Ti。在被赶走的过程中牛就不能乱吃了

要让被吃掉的花朵数量最少，我想应该是跟牛所在的位置和吃花的速度有关，于是尝试了一下按照a.D[i] * b.T[i] > a.T[i] * b.D[i] 排序，没问题。

#include <iostream>
#include<algorithm>
#include<queue>
#include<cstdio>
#include<cmath>
#include<cstring>

using namespace std;
const int MAXN = 2000020;
typedef long long LL;
int n;
typedef pair<LL,LL> P;
P node[MAXN];
bool cmp(P a,P b)
{
    return a.second * b.first > b.second * a.first;
}
int main()
{
   // freopen("in.txt","r",stdin);
    LL sum = 0;
    scanf("%d",&n);
    for(int i = 0;i<n;i++)
    {
        scanf("%I64d%I64d",&node[i].first,&node[i].second);
        sum += node[i].second;
    }
    sort(node,node+n,cmp);
    LL ans = 0;
    for(int i = 0;i<n;i++)
    {
        sum -= node[i].second;
        ans += sum*(node[i].first*2);
    }
    printf("%I64d\n",ans);
    return 0;
}