POJ 2631 Roads in the North【裸题】

Roads in the North Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2628   Accepted: 1298 Description Building and maintaining roads among communities in the far North is an expensive business. With this in mind, the roads are build such that there is only one route from a village to a village that does not pass through some other village twice.  Given is an area in the far North comprising a number of villages and roads among them such that any village can be reached by road from any other village. Your job is to find the road distance between the two most remote villages in the area.  The area has up to 10,000 villages connected by road segments. The villages are numbered from 1.  Input Input to the problem is a sequence of lines, each containing three positive integers: the number of a village, the number of a different village, and the length of the road segment connecting the villages in kilometers. All road segments are two-way. Output You are to output a single integer: the road distance between the two most remote villages in the area. Sample Input 5 1 6 1 4 5 6 3 9 2 6 8 6 1 7 Sample Output 22 Source The UofA Local 1999.10.16

这是道树的直径的裸题，大意是给了各个村庄之间的距离，求离得最远的两个村庄之间的距离。

贴一下模板，方便查询吧。

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;

#define N 10000+10
struct a
{
int t,w,next;
}edge[N*2];
int head
,vis
,dis
;
int ans,Tnode,edgenum;
void addedge(int u,int v,int w)//储存该图
{
edge[edgenum].t=v;
edge[edgenum].w=w;
edge[edgenum].next=head[u];
head[u]=edgenum++;
}

void bfs(int s)//用bfs对图进行搜索
{
int i,u,v;
memset(vis,0,sizeof(vis));
memset(dis,0,sizeof(dis));
queue<int> q;
q.push(s);
vis[s]=1;dis[s]=0;ans=0;
while(!q.empty())
{
u=q.front();
q.pop();
for(i=head[u];i!=-1;i=edge[i].next)
{
v=edge[i].t;
if(!vis[v])
{
if(dis[v]<dis[u]+edge[i].w)
{
dis[v]=dis[u]+edge[i].w;
if(ans<dis[v])
{
ans=dis[v];
Tnode=v;
}
}
vis[v]=1;
q.push(v);
}
}
}
}
int main()
{
int x,y,w;
memset(head,-1,sizeof(head));
edgenum=0;
while(~scanf("%d%d%d",&x,&y,&w))
{
addedge(x,y,w);
addedge(y,x,w);
}
bfs(1);//第一次找起点
bfs(Tnode);//第二次从起点开始找终点
printf("%d\n",ans);
return 0;
}