POJ3262 Protecting the Flowers 【贪心】

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 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

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.

Source

USACO 2007 January Silver

被long long坑了6次...贪心方法为每次选择的牛使得剩下的牛对花园破坏最小。

证明：若a,b为即将选择的两头牛，若果先选a的话。b牛的破坏是b.d*a.t,若先选b的话。a的破坏是a.d*b.t;无论a,b谁先谁后，其余牛的破坏都是(a.t+b.t)*other.d,所以仅仅须要依照先选择的牛使得剩下的牛对花园的破坏最小的标准对牛排序，再逐个选择就是了。

#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>

#define maxn 100005
typedef 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;
}