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How far away ？
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7090 Accepted Submission(s): 2578
Problem Description
There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this
village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can‘t visit a place twice) between every two houses. Yout task is to answer all these curious people.
Input
First line is a single integer T(T<=10), indicating the number of test cases.
For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road
connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.
Output
For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.
Sample Input
2
3 2
1 2 10
3 1 15
1 2
2 3
2 2
1 2 100
1 2
2 1
Sample Output
Source
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ac代码
#include<stdio.h>
#include<string.h>
int vis[40040],head[40040],n,m,x[40040],y[40040],dis[40040],cnt,pre[40040],z[40040];
struct s
{
int u,v,w,next;
}edge[40005*2];
void init()
{
memset(vis,0,sizeof(vis));
memset(head,-1,sizeof(head));
memset(dis,0,sizeof(dis));
cnt=0;
}
void add(int u,int v,int w)
{
edge[cnt].u=u;
edge[cnt].v=v;
edge[cnt].w=w;
edge[cnt].next=head[u];
head[u]=cnt++;
}
int find(int x)
{
if(x==pre[x])
return x;
return find(pre[x]);
}
void tarjan(int u)
{
int i;
pre[u]=u;
vis[u]=1;
for(i=1;i<=m;i++)
{
if(x[i]==u&&vis[y[i]])
z[i]=find(y[i]);
else
if(y[i]==u&&vis[x[i]])
z[i]=find(x[i]);
}
for(i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(!vis[v])
{
dis[v]=dis[u]+edge[i].w;
tarjan(v);
pre[v]=u;
}
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
int i;
init();
for(i=0;i<n-1;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
add(v,u,w);
}
for(i=1;i<=m;i++)
{
scanf("%d%d",&x[i],&y[i]);
}
tarjan(1);
for(i=1;i<=m;i++)
{
printf("%d\n",dis[x[i]]+dis[y[i]]-2*dis[z[i]]);
}
}
}
HDOJ 题目2586 How far away ？（LCA）
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原文地址：http://blog.csdn.net/yu_ch_sh/article/details/45099811
