Silver Cow Party （dijkstra）

最新推荐文章于 2021-08-05 18:17:02 发布

原创最新推荐文章于 2021-08-05 18:17:02 发布 · 373 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#acm #dijkstra

本文介绍了一个关于牛从各自农场前往聚会地点再返回的路径选择问题。每头牛都会选择最短路径，需找出所有牛中最长往返时间。采用两次Dijkstra算法解决此问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A - Silver Cow Party

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 T_i (1 ≤ T_i ≤ 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: A_i, B_i, and T_i. The described road runs from farm A_i to farm B_i, requiring T_i 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.

题目大意：有很多牛要从自己的农场去指定的农场参加聚会，然后再返回自己所在的农场，来回的路程是不一样的，但是牛都会优先走最短的，求所有牛中走的最长的路程是多少

他们都说什么转换数组，我不懂，我用的两个 dijkstra解决的，代码比较傻，但是容易理解。

#include<stdio.h>
#include<string.h>
using namespace std;
#define maxn 1010
#define inf 99999999
int n;
int e[maxn][maxn];
int dis1[maxn],dis2[maxn];
void d1(int x)
{
	int i,j,min,u,v;
	int book[1010];
	for(i=1;i<=n;i++)
		dis1[i]=e[i][x];
	memset(book,0,sizeof(book));
	book[x]=1;
	for(i=1;i<=n-1;i++)
	{
		min=inf;
		for(j=1;j<=n;j++)
		{
			if(book[j]==0 && dis1[j]<min)
			{
				u=j;
				min=dis1[j];
			}
		}
		book[u]=1;
		for(v=1;v<=n;v++)
		{
			if(e[v][u]<inf)
			{
				if(dis1[v]>dis1[u]+e[v][u])
					dis1[v]=dis1[u]+e[v][u];
			}
		}
	}
}
void d2(int x)
{
	int i,j,min,u,v;
	int book[1010];
	for(i=1;i<=n;i++)
		dis2[i]=e[x][i];
	memset(book,0,sizeof(book));
	book[x]=1;
	for(i=1;i<=n-1;i++)
	{
		min=inf;
		for(j=1;j<=n;j++)
		{
			if(book[j]==0 && dis2[j]<min)
			{
				u=j;
				min=dis2[j];
			}
		}
		book[u]=1;
		for(v=1;v<=n;v++)
		{
			if(e[u][v]<inf)
			{
				if(dis2[v]>dis2[u]+e[u][v])
					dis2[v]=dis2[u]+e[u][v];
			}
		}
	}
}
int main()
{
	//printf("%d",inf);
	int m,x,i,j,a,b,t;
	while(scanf("%d%d%d",&n,&m,&x)!=EOF)
	{
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			{
				if(i==j)
					e[i][j]=0;
				else
					e[i][j]=inf;
			}
		}
		while(m--)
		{
			scanf("%d%d%d",&a,&b,&t);
			if(e[a][b]>t)
				e[a][b]=t;
		}
		d1(x);
		d2(x);
		int ans=0;
		for(i=1;i<=n;i++)
		{
			if(dis1[i]+dis2[i]>ans)
				ans=dis1[i]+dis2[i];
			//printf("%d到%d为%d----%d到%d为%d\n",i,x,dis1[i],x,i,dis2[i]);
		}
		printf("%d\n",ans);
	}
}

注意d1和d2的微小差别就不会错！