POJ1985 Cow Marathon

最新推荐文章于 2020-06-07 12:23:26 发布

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本文介绍了一种使用两次广度优先搜索(BFS)算法来解决寻找地图上两个最远农场间最大距离的问题。通过构建图并应用BFS，可以有效地找到最长路径。

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Cow Marathon

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 5054		Accepted: 2459
Case Time Limit: 1000MS

Description

After hearing about the epidemic of obesity in the USA, Farmer John wants his cows to get more exercise, so he has committed to create a bovine marathon for his cows to run. The marathon route will include a pair of farms and a path comprised of a sequence of roads between them. Since FJ wants the cows to get as much exercise as possible he wants to find the two farms on his map that are the farthest apart from each other (distance being measured in terms of total length of road on the path between the two farms). Help him determine the distances between this farthest pair of farms.

Input

* Lines 1.....: Same input format as "Navigation Nightmare".

Output

* Line 1: An integer giving the distance between the farthest pair of farms.

Sample Input

Sample Output

Hint

The longest marathon runs from farm 2 via roads 4, 1, 6 and 3 to farm 5 and is of length 20+3+13+9+7=52.

Source

USACO 2004 February

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题意：

给出各农场之间的距离，求两个农场之间最大的距离。用两次BFS就可以了。

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int maxn=80010;
int n,m;
int head[maxn];
struct node{
	int v,w,next;
}edge[maxn];
int cnt=0;
void addedge(int u,int v,int w)
{
	edge[cnt].v=v;
	edge[cnt].w=w;
	edge[cnt].next =head[u];
	head[u]=cnt++;
}
void fun()
{
	memset(head,-1,sizeof(head));
}
int vis[maxn],dis[maxn],q[maxn],res;
int bfs(int u)
{
	memset(vis,0,sizeof(vis));
	int front=0;
	int rear=1;
	q[1]=u;
	vis[u]=1;
	dis[u]=0;
	int ans=0;
	while(front<rear)
	{
		front++;
		int tmp=q[front];
		for(int i=head[tmp];i!=-1;i=edge[i].next)
		{
			int v=edge[i].v;
			if(!vis[v])
			{
				rear++;
				q[rear]=v;
				vis[v]=1;
				dis[v]=dis[tmp]+edge[i].w ;
				if(dis[v]>ans)
				{
					ans=dis[v];
					res=v;
				}
			}
		}
	}
	return ans;
}
int main()
{
	int u,v,w;
	char ch;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		fun();
		cnt=0;
		for(int i=0;i<m;i++)
		{
			scanf("%d %d %d %c",&u,&v,&w,&ch);
			addedge(u,v,w);
			addedge(v,u,w);
		}
		bfs(1);
		printf("%d\n",bfs(res));
	}
	return 0;
}