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[112345];
int N,M,total,head[112345],dis[112345];
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[112345]; int num,sta[112345],tab[112345],dep[112345],dp[112345][22];
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 != -1; i = edge[i].next ){
       int v = edge[i].v; //cout<<edge[i].val<<endl;
       if( !vis[v] )
       { LCA( v,deep+1,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 = 1; i <= num; i++ )dp[i][0] = i;
    for( int i = 1; (1<<i) <= num; i++ )
    for( int j = 1; j - 1 + (1<<i) <= num; j++ )
        dp[j][i] = work( dp[j][i-1],dp[j+(1<<(i-1))][i-1] );
}
int query( int L,int R )
{
    int k = 0;
    while( (1<<(k+1)) <= (R-L+1) )k++;
    return tab[work( dp[L][k],dp[R-(1<<k)+1][k] )];
}
int main( )
{
    int u,v,w; char str[3];
    while( scanf("%d%d",&N,&M) != EOF )
    {
         memset( head,-1,sizeof(head) ); total = 0;
         for( int i = 1; i <= M; i++ )
         {
            scanf("%d%d%d%s",&u,&v,&w,&str);
            add_edge( u,v,w );
            add_edge( v,u,w );
         }
         memset( vis,0,sizeof(vis) );
         num = 1; LCA( 1,1,0 ); 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]-2*dis[t]<<endl;
         }
    }
    return 0;
}