POJ3262 Protecting the Flowers (贪心)

Protecting the Flowers

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 7941 Accepted: 3201

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 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

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

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.

Source

USACO 2007 January Silver

题目链接：http://poj.org/problem?id=3262

题目大意：有n个牛在FJ的花园吃花，FJ要把它们赶回去，每次只能赶回去一头牛，耗费的时间为Ti * 2，某头牛在被赶回去之前破坏的花的数量为每分钟Di，求FJ把所有牛都赶回去并且使得被破坏的花的数量最少。

思路：是存在一个序列.......(Ti,Di)(Ti+1,Di+1)，，，，FJ赶牛花费的时间为X，则要把第i头和第i+1头牛全部赶回去要花费的代价为：X*Di + (X + Ti) * Di+1。

把(Ti,Di)和(Ti+1,Di+1)进行交换，则要花费的代价为：X*Di+1 + (X + Ti+1) * Di；

两式做差可得Ti * Di+1 - Ti+1 * Di   。假设两者之差小于0，则证明交换后花费的代价较大，Ti * Di+1 - Ti+1 * Di < 0 ，则Ti * Di+1 < Ti+1 * Di。

AC代码如下：

#include <cstdio>#include <cstring>#include <algorithm>#define ll long longusing namespace std;const int maxn = 1e5 + 5;struct node{int t,d;}p[maxn];int n;bool cmp(node A, node B){return A.d*1.0 / A.t > B.d*1.0 / B.t;}int main(){while(~scanf("%d",&n)){ll ans = 0;for(int i = 0; i < n; i ++){scanf("%d%d",&p[i].t,&p[i].d);ans += p[i].d;}sort(p,p+n,cmp);ll sum = 0;for(int i = 0; i < n; i ++){ans -= p[i].d;sum += p[i].t * 2 *ans;}printf("%lld\n",sum);}return 0;}