POJ 3262 - Protecting the Flowers（贪心）

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 andD_i, 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

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思路：

猜是和 d/t 有关的，可以证明得到。

考虑当前选哪个放在开头。
设放在开头的为(d1,t1) 放在第二个的为(d2,t2)。
那么放在开头的牛满足：无论第二个是什么，交换第一和第二总是不会变优。
设当前剩余牛的d值和为ds。
那么有：
(ds-d1)*t1+(ds-d1-d2)*t2<(ds-d2)*t2+(ds-d2-d1)*t1
化简得：
 t2/d2>t1/d1。
也就是说，当t[i]/d[i]>t[j]/t[j]时，j放在i前面更优。

CODE：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 100005;typedef long long ll;struct node {    ll t, d;    double b;    bool operator < (const node &x) const {        return x.b < b;    }}num[maxn];ll des[maxn];int main(){//freopen("in", "r", stdin);    int n;    while(~scanf("%d", &n)) {        for(int i = 0; i < n; ++i) {            scanf("%I64d %I64d", &num[i].t, &num[i].d);            num[i].b = (double)num[i].d / num[i].t;        }        sort(num, num + n);        des[0] = num[0].d;        for(int i = 1; i < n; ++i) {            des[i] = des[i-1] + num[i].d;        }        ll ans = 0;        for(int i = 0; i < n; ++i) {            ans +=  num[i].t * 2 * (des[n-1]-des[i]);        }        printf("%I64d\n", ans);    }    return 0;}