poj 3262 Protecting the Flowers

最新推荐文章于 2021-01-30 04:05:14 发布

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本文探讨了一个经典的优化问题，即如何通过合理的调度顺序来最小化牛群在被运送回各自谷仓过程中破坏的花朵数量。该问题通过算法解决，重点在于通过计算每头牛破坏花朵速率和距离其谷仓的时间来确定最优运输顺序。

Protecting the Flowers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4544		Accepted: 1820

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

4 11 6

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

注意定义sum_J和sum时要用长整形，一直就错在这里

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAX 100000
__int64 sum_J[MAX+10],sum;
struct Cow
{
	int T;
	int D;
	double P;
}num[MAX+10];
int cmp(const void *a,const void *b)
{
	struct Cow *c=(Cow *)a;
	struct Cow *d=(Cow *)b;
	if(c->P!=d->P)
	return d->P>c->P?1:-1;
}
int main()
{
	int N,i,j,k;
	while(scanf("%d\n",&N)!=EOF)
	{
		for(i=0;i<N;i++)
		{
			scanf("%d %d",&num[i].T,&num[i].D);
			num[i].P=(num[i].D*1.0)/(num[i].T*1.0);
		}
		qsort(num,N,sizeof(num[0]),cmp);
		
		memset(sum_J,0,sizeof(sum_J));
		for(i=N-2,sum_J[N-1]=num[N-1].D;i>=0;i--)
		sum_J[i]=sum_J[i+1]+num[i].D;
//		for(i=0;i<N;i++)
//		printf("%d#\n",sum_J[i]);
		
		for(i=0,sum=0;i<N-1;i++)
		{
//			for(j=i+1,S=0;j<N;j++)
//			S+=num[j].D;
			
			sum+=(sum_J[i+1]*num[i].T);
		}
		printf("%I64d\n",2*sum);//前边定义sum要用长整形，否则会错。（这里一直错） 
		//for(i=0;i<N;i++)
		//printf("%d %d %lf\n",num[i].T,num[i].D,num[i].P);
		
	}
	return 0;
}