poj 1986 Distance Queries

好像是模板题  当作练习题 不错；  要求任意两点之间的距离。可以假设一个根节点，然后所有点到根节点的距离，然后求出任意两点多公共祖先；  距离就变成了

dis[u]+dis[v] - 2*dis[ lca(u,v) ]  非常好的题目

 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;

 struct date{
    int v,val,next;
 }edge[];
 int N,M,total,head[],dis[];
 void add_edge( int u,int v,int val ){
    edge[total].v = v;
    edge[total].val = val;
    edge[total].next = head[u];
    head[u] = total++;
 }
 bool vis[]; int num,sta[],tab[],dep[],dp[][];
 void LCA( int son,int deep,int w ){
     dep[num] = deep; tab[num] = son; sta[son] = num++; dis[son] = w; vis[son] = true;
     for( int i = head[son]; i != -; i = edge[i].next ){
        int v = edge[i].v; //cout<<edge[i].val<<endl;
        if( !vis[v] )
        { LCA( v,deep+,w+edge[i].val ); dep[num] = deep; tab[num] = son; num++; }
     }
     //dep[num] = deep; tab[num] = son; num++;
 }
 int work( int n1,int n2 ){
    if( dep[n1] < dep[n2] )return n1;
    return n2;
 }
 void RMQ( ){
     for( int i = ; i <= num; i++ )dp[i][] = i;
     for( int i = ; (<<i) <= num; i++ )
     for( int j = ; j -  + (<<i) <= num; j++ )
         dp[j][i] = work( dp[j][i-],dp[j+(<<(i-))][i-] );
 }
 int query( int L,int R )
 {
     int k = ;
     while( (<<(k+)) <= (R-L+) )k++;
     return tab[work( dp[L][k],dp[R-(<<k)+][k] )];
 }
 int main( )
 {
     int u,v,w; char str[];
     while( scanf("%d%d",&N,&M) != EOF )
     {
          memset( head,-,sizeof(head) ); total = ;
          for( int i = ; i <= M; i++ )
          {
             scanf("%d%d%d%s",&u,&v,&w,&str);
             add_edge( u,v,w );
             add_edge( v,u,w );
          }
          memset( vis,,sizeof(vis) );
          num = ; LCA( ,, ); num--; RMQ( );
          //for( int i = 1; i <=  num; i++ )cout<<i<<" "<<dep[i]<<" "<<tab[i]<<endl;
          int Q; scanf("%d",&Q);
          while( Q-- ){
             scanf("%d%d",&u,&v);
             //cout<<u<<" "<<v<<" "<<dis[u]<<" "<<dis[v]<<endl;
             if( sta[u] > sta[v] )swap( u,v );
             int t = query(sta[u],sta[v]);//cout<<endl<<" "<<t<<endl;
             //cout<<t<<" "<<dis[1]<<" "<<u<<" "<<v<<" "<<dis[u]<<" "<<dis[v]<<endl;
             cout<<dis[u]+dis[v]-*dis[t]<<endl;
          }
     }
     return ;
 }