POJ 3268 - Silver Cow Party （dijkstra）

最新推荐文章于 2021-04-01 18:48:34 发布

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本文介绍了一道关于寻找N个农场中牛群前往特定农场参加派对并返回的最长时间路径问题。通过两次使用Dijkstra算法计算双向最短路径，解决了Floyd算法超时的问题。

One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires Ti (1 ≤ Ti ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input

Line 1: Three space-separated integers, respectively: N, M, and X
Lines 2.. M+1: Line i+1 describes road i with three space-separated integers: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

Sample Input

Sample Output

Hint

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

题意：

各个农场的牛要去X农场参加派对，求他们往返路程的最大值。

思路：

用两次dijkstra求出双向的距离相加就行了。

一开始用的Floyd结果超时了，还是用dijkstra比较好。

代码：

#include<stdio.h>
#include<string.h>
#define inf 99999999
int map[1010][1010],dis1[1010],dis2[1010],book[1010];
int main()
{
	int n,m,x,a,b,c,i,j,k,min,max,u;
	while(scanf("%d%d%d",&n,&m,&x)!=EOF)
	{
		for(i=1;i<=n;i++)
			for(j=1;j<=n;j++)
			{
				if(i==j)
					map[i][j]=0;
				else
					map[i][j]=inf;
			}
		for(i=1;i<=m;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			map[a][b]=c;
		}
		for(i=1;i<=n;i++)
			dis1[i]=map[x][i];
		for(i=1;i<=n;i++)
			dis2[i]=map[i][x];
		memset(book,0,sizeof(book));
		for(i=1;i<n;i++)
		{
			min=inf;
			for(j=1;j<=n;j++)
			{
				if(book[j]==0&&dis1[j]<min)
				{
					min=dis1[j];
					u=j;
				}
			}
			book[u]=1;
			for(k=1;k<=n;k++)
			{
				if(book[k]==0&&dis1[k]>dis1[u]+map[u][k])
					dis1[k]=dis1[u]+map[u][k];
			}
		}
		memset(book,0,sizeof(book));
		for(i=1;i<n;i++)
		{
			min=inf;
			for(j=1;j<=n;j++)
			{
				if(book[j]==0&&dis2[j]<min)
				{
					min=dis2[j];
					u=j;
				}
			}
			book[u]=1;
			for(k=1;k<=n;k++)
			{
				if(book[k]==0&&dis2[k]>dis2[u]+map[k][u])
					dis2[k]=dis2[u]+map[k][u];
			}
		}
		max=0;
		for(i=1;i<=n;i++)
		{
			if(dis1[i]+dis2[i]>max)
				max=dis1[i]+dis2[i];
		}
		printf("%d\n",max);
	}
	return 0;
}