Poj 3262 Protecting the flowers

最新推荐文章于 2019-06-05 21:46:24 发布

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最新推荐文章于 2019-06-05 21:46:24 发布

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分类专栏：基础算法-分治-递归-枚举-贪心－模拟。。。

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本文探讨了一种方法，通过合理安排奶牛返回牧场的顺序，以最小化花园中花卉的破坏。利用数学模型计算得出最优路径，确保在奶牛返回各自牧场的过程中，减少对花园花卉的影响。

Protecting the Flowers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3629		Accepted: 1490

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

Sample Output

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.

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设放在开头的为(d1,t1) 放在第二个的为(d2,t2)。
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。
所以我们可以按t[i]/d[i]排序，从小取到大就是最终序列了。

#include<stdio.h>
#include<algorithm>
using namespace std;

struct node{
    __int64 T;
    __int64 D;
}Node[100010];
__int64 sum[100010];

bool cmp(struct node a,struct node b){
    return (double)a.T/a.D < (double)b.T/b.D;
}

int main(){
    int n;
    int i;
    __int64 mins;
    while(scanf("%d",&n)!=EOF){
        mins = 0;
        for(i = 0;i<n;i++) {
            scanf("%I64d%I64d",&Node[i].T,&Node[i].D);
            Node[i].T*=2;
        }
        sort(Node,Node+n,cmp);
        sum[n-1] = Node[n-1].D;
        for(i = n-2;i>=0;i--) sum[i] = sum[i+1]+Node[i].D;
        for(i = 1;i<n;i++) mins += sum[i]*Node[i-1].T;
        printf("%I64d\n",mins);
    }
    return 0;
}