POJ 3262 -- Protecting the Flowers

http://poj.org/problem?id=3262

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

Sample Input

6
3 1
2 5
2 3
3 2
4 1
1 6

http://www.hankcs.com/program/cpp/poj-3262-protecting-the-flowers.html

约翰的奶牛每分钟吃掉D_i朵花，把它赶走需要T_i分钟（来回加倍）。问最小损失花朵数量。

贪心策略是尽量赶走吃得多并且走得慢的牛，如何衡量“多”“慢”？需要两头牛作比较:

1 2 3 4

bool operator < (const Cow& c) const     {         return T * c.D < c.T * D;     }

struct Cow

{

    int T;

    int D;

    bool operator < (const Cow& c) const

    {

        return T * c.D < c.T * D;

    }

};

 

Cow cow[100000];

 

int main(int argc, char *argv[])

{

#ifndef ONLINE_JUDGE

    freopen("in.txt", "r", stdin);

    freopen("out.txt", "w", stdout);

#endif

    int N;

    cin >> N;

    int total_destory = 0;

    for (int i = 0; i < N; ++i)

    {

        cin >> cow[i].T >> cow[i].D;

        total_destory += cow[i].D; // 总的损害

    }

    sort(cow, cow + N);

    unsigned long long destroied = 0;

    for (int i = 0; i < N; ++i)

    {

        total_destory -= cow[i].D;

        destroied += total_destory * cow[i].T * 2; // 损害维持时间为2倍的 moving 牛耗时

    }

    cout << destroied << endl;

#ifndef ONLINE_JUDGE

    fclose(stdin);

    fclose(stdout);

    system("out.txt");

#endif

    return 0;

}

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