Protecting the Flowers

小灰灰 2022-06-12 06:29 107阅读 0赞

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

题意：有n头牛在破坏庄稼,每头牛拉回家的时间t和每分钟破坏庄稼的速度d,

求怎样将牛拉回家才使得破坏的庄稼最少，每次只能拉一头牛回家

#include<cstdio>
    #include<algorithm>
    using namespace std;
    struct A{
      __int64 a,b;
    }f[100000];
    bool cmp(A x,A y)
    {
        return x.b*1.0/x.a>y.b*1.0/y.a;//因为破坏的庄稼跟时间,破坏速度都有关,想要破坏的庄稼最少，
        //当然就是破坏速度越慢越好,这个好理解,难以理解的是时间,这个比值的意思可以理解为:破坏速度快的和拉牛回家的时间少的排在前面
    }
    int main()
    {
        int n,i;
        __int64 s=0,ans=0;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%I64d %I64d",&f[i].a,&f[i].b);
            s=s+f[i].b;//s是所有的牛每分钟破坏庄稼的速度
        }
        sort(f,f+n,cmp);
        for(i=0;i<n;i++)
        {
            //printf("%I64d %I64d\n",f[i].a,f[i].b);
            s=s-f[i].b;
            ans+=f[i].a*2*s;
        }
        printf("%I64d\n",ans);
        return 0;
    }

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.