[POJ3262]Protecting the Flowers

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.
假设只有两头牛a, b 
若先取走a牛, 则食花量为2 * a.t * b.d
若先取走b牛, 则食花量为2 * b.t * a.d
两式分别除以a.d * b.d, 得到2 * a.t / a.d,  2 * b.t / b.d

题意：N头奶牛在农夫约翰的花园里吃花。他必须将奶牛赶回它自己的牛栏。从花园到牛栏i需要Ti(1<=Ti<=2000000)分钟的时间，然后回到花园又需要Ti分钟。第i头奶牛每分钟吃掉Di朵花（1<=Di<=100）。赶牛的过程中，奶牛就不能吃花了。问农夫约翰如何赶牛才能使得花园损失最小。求最小的损失。

题解：t/d值小的优先

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>using namespace std;typedef long long LL;const int N=1e5+10;struct node{ LL t, d; double m; }cow[N];LL n, sum;bool cmp( node a, node b ) {return a.m<b.m;}int main() {scanf( "%I64d", &n );for( int i=1; i<=n; i++ ) {scanf( "%I64d%I64d", &cow[i].t, &cow[i].d );double t=cow[i].t, d=cow[i].d;cow[i].m=t/d;}sort( cow+1, cow+n+1, cmp );for( int i=1; i<=n; i++ ) cow[i].d+=cow[i-1].d;for( int i=1; i<=n; i++ )sum+=2*cow[i].t*( cow[n].d-cow[i].d );printf( "%I64d\n", sum );return 0;}