POJ3262 Protecting the Flowers 【貪心】，poj3262protecting

POJ3262 Protecting the Flowers 【貪心】，poj3262protecting

Protecting the Flowers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4418		Accepted: 1785

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

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被long long坑了6次...貪心方法為每次選擇的牛使得剩下的牛對花園破壞最小。證明：若a,b為即將選擇的兩頭牛，若果先選a的話，b牛的破壞是b.d*a.t,若先選b的話，a的破壞是a.d*b.t;不管a,b誰先誰後，其餘牛的破壞都是(a.t+b.t)*other.d,所以只需要按照先選擇的牛使得剩下的牛對花園的破壞最小的標準對牛排序，再逐個選擇就是了。

#include <stdio.h>#include <string.h>#include <queue>#include <algorithm>#define maxn 100005typedef long long LL;struct Node {    int t, d;    friend bool operator<(const Node& a, const Node& b) {        return a.t * b.d < b.t * a.d;    }} cow[maxn];int main() {    int N, i, sumt = 0;    LL sumd = 0;    scanf("%d", &N);    for(i = 0; i < N; ++i)        scanf("%d%d", &cow[i].t, &cow[i].d);    std::sort(cow, cow + N);    for(i = 0; i < N; ++i) {        sumd += cow[i].d * sumt;        sumt += cow[i].t * 2;    }    printf("%lld\n", sumd);    return 0;}