Protecting the Flowers (贪心)

最新推荐文章于 2020-08-16 16:23:21 发布

爱上键盘的小哥哥

最新推荐文章于 2020-08-16 16:23:21 发布

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分类专栏：贪心算法

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面对一群在花园中破坏花朵的牛，农夫约翰必须制定一个最优策略将它们送回各自的畜舍，以最小化花的损失。通过合理安排运送顺序，可以有效地减少破坏。

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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 destroysD_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 cowi to its barn requires 2 × T_i minutes (T_i to get there andT_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 andD_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.、

题意：

有N头牛在花园里破坏花，因此老汉要把牛全部运会圈里，运回每头牛的所花时间不同，分别为Ti，每头牛每分钟破坏花的数量为Di，求把所有牛运回，使花破坏最少的数量

思路：

假设有两头牛 A1，A2，运回A1 花时间 T1，每分钟破坏花的数量为D1；运回A2 花时间 T2，每分钟破坏花的数量为D2

如果先运回A1 破坏花的数量为 2*T1*D2

如果先运回A2 破坏花的数量为 2*T2*D1

两式同时除以2* D1*D2 则为T1/D1 T2/D2

所以对于每头牛 T/D 越小越先运回

代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node{
    int d;
    int t;
    double num;
}cow[100050];
int cmp(node a,node b){
    return a.num<b.num;
}
int main(){
    int n;
    scanf("%d",&n);
    int i;
    int D=0;
    for(i=0;i<n;i++){
        scanf("%d%d",&cow[i].t,&cow[i].d);
        cow[i].num=cow[i].t/(cow[i].d*1.0);
       // printf("%.3f\n",cow[i].num);
        D+=cow[i].d;
    }
   // printf("D %d\n",D);
    sort(cow,cow+n,cmp);
   // D-=cow[n-1].d;
    long long  sum=0;
    for(i=0;i<n;i++){

        D-=cow[i].d;  //牛被运走 对花无破坏
        sum+=cow[i].t*2*D; //剩下所有牛每分钟破坏花的数量  乘以 运牛来回所花的时间
    }
    printf("%I64d\n",sum);
return 0;
}