POJ：3262-Protecting the Flowers

Protecting the Flowers

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 8606 Accepted: 3476

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.

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解题心得：

1. 题意就是有n头牛在吃花，每头牛牵走（避免吃花）要花费2*t秒（来回），每头牛吃花的速率是d，要你自己设计一个牵走牛的顺序，使被牛吃的花最少。
2. 假设有两头牛分别编号1,2，1号牛牵走需要花费x1秒，吃花的速率是y1，2号牛牵走需要x2秒，吃花的速率是y2，假设先牵走1号牛，那么2号牛将要吃y2*x1数量的花，如果先牵走2号牛，那么1号牛要吃y1*x2数量的花，假设二号牛吃得花比1号牛多，那么就有关系式y2*x1>y1*x2，那么可以移项得到y2/x2>x2/x1，所以这样就可以按照这个速率比时间的值来贪心，让被牛吃的花尽量少。

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#include <stdio.h>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 1e5+100;
struct NODE {
    ll d,t;
    double v;
}cow[maxn];

bool cmp(NODE a, NODE b) {
    return a.v < b.v;
}

int main() {
    ll n,t = 0,ans = 0;
    scanf("%lld",&n);
    for(int i=0;i<n;i++){
        scanf("%lld%lld",&cow[i].t,&cow[i].d);
        cow[i].v = (double)cow[i].t/(double)cow[i].d;
    }
    sort(cow,cow+n,cmp);
    for(int i=0;i<n;i++) {
        ans += cow[i].d * t;
        t += cow[i].t*2;
    }

    printf("%lld
",ans);
    return 0;
}