http://acm.hdu.edu.cn/showproblem.php?pid=2586&&How far away ？

[align=left]Problem Description[/align]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 thisvillage 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.[align=left]Input[/align]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 roadconnecting 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.[align=left]Output[/align]For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.[align=left]Sample Input[/align]

23 21 2 103 1 151 22 32 21 2 1001 22 1[align=left]Sample Output[/align]

1025100100题意:最短路问题，Tarjan离线算法。。。AC代码：

#include<cstdio>
#include<string.h>
#include<algorithm>
#include<string>
#include<iostream>
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FOR(s,t) for(int s=0;i<t;++i)
#define N 40005
#define M 81000
using namespace std;
typedef struct
{
int num;
int len;
int Next;
}Node;
Node s[M];
int head1
,head
,Father
,id
,dis
,ans[405];
bool vis
;
int res,cnt;
void init()
{
CLR(head,-1);
CLR(head1,-1);
CLR(id,0);
CLR(ans,0);
CLR(vis,false);
res=0,cnt=0;
}
int Find(int x)
{
if(x==Father[x]) return x;
return Father[x]=Find(Father[x]);
}
void add(int a,int b,int c,int h[])
{
s[res].num=b,s[res].len=c,s[res].Next=h[a],h[a]=res++;
s[res].num=a,s[res].len=c,s[res].Next=h[b],h[b]=res++;
}
void Tarjan(int x)
{
id[x]=cnt;
vis[Father[x]=x]=true;
int v;
for(int i=head1[x];i!=-1;i=s[i].Next)
if(vis[v=s[i].num])
if(id[x]==id[v]) ans[s[i].len]=dis[x]+dis[v]-2*dis[Find(v)];
for(int i=head[x];i!=-1;i=s[i].Next)
if(!vis[v=s[i].num])
dis[v]=dis[x]+s[i].len,Tarjan(v),Father[v]=x;
}
void in(int &a)
{
char ch;
while((ch=getchar())<'0'||ch>'9');
for(a=0;ch>='0'&&ch<='9';ch=getchar()) a=a*10+ch-'0';
}
int main()
{
int T;
in(T);
for(int k=1;k<=T;++k)
{
if(k>1) printf("\n");
init();
int n,m,a,b,c;
in(n),in(m);
FOR(i,n-1) in(a),in(b),in(c),add(a,b,c,head);
FOR(i,m) in(a),in(b),add(a,b,i,head1);
for(int i=1;i<=n;++i,cnt++)
if(!vis[i]) dis[i]=0,Tarjan(i);
FOR(i,m) printf("%d\n",ans[i]);
}return 0;
}