poj 1986 Distance Queries (LCA)

题目链接:   poj 1986

题目大意:   给出一棵树,求a—b路径的长度

解题思路:   与hdu 2874类似

                  LCA离线查找最近公共祖先

                  dist[ ]求距离: 距离=dist[ a ] + dist[ b ] — 2*dist[ LCA(a,b) ]

代码:

//Final LCA离线 查找两点最短距离#include <stdio.h>#include <string.h>#include <stdlib.h>#define MAX 50100struct {int to,w,next;}Hash[MAX<<2],Qes[MAX<<2];  //Hash是存在的边，Qes是需要查询点int n,Index1,Index2,pre1[MAX],pre2[MAX],visit[MAX],ansetor[MAX],ans[MAX],dist[MAX];void Add_edge(int a,int b,int c)   //建立树{Hash[Index1].to=b,Hash[Index1].w=c;Hash[Index1].next=pre1[a];pre1[a]=Index1++;}void Add_Qes(int a,int b,int c)    //建立离线查询{Qes[Index2].to=b,Qes[Index2].w=c;Qes[Index2].next=pre2[a];pre2[a]=Index2++;}int Find(int x)           //最近公共祖先查找{int s,j;s=x;while(x!=ansetor[x])x=ansetor[x];while(s!=x){j=ansetor[s];ansetor[s]=x;s=j;}return x;}void LCA(int u)     //离线查找最近公共祖先，并且在线更新当前点到根结点的距离{    int i,v;    visit[u]=1;    ansetor[u]=u;    for(i=pre1[u];i!=-1;i=Hash[i].next)    {        v=Hash[i].to;        if(!visit[v])        //如果此点没有访问过，则更新dist[]        {            dist[v]=dist[u]+Hash[i].w;            LCA(v);          //回溯            ansetor[v]=u;    //当前点的祖先暂时是u        }    }    for(i=pre2[u];i!=-1;i=Qes[i].next)    {        v=Qes[i].to;        if(visit[v])          //如果这个点之前访问过，那么它最近的祖先也就确定了        {            ans[Qes[i].w]=dist[u]+dist[v]-dist[ansetor[Find(v)]]*2;        }    }}int main(){char ch[2];int m,k,i,a,b,c;while(scanf("%d%d",&n,&m)!=EOF){Index1=Index2=0;memset(ans,-1,sizeof(ans));memset(pre1,-1,sizeof(pre1));memset(pre2,-1,sizeof(pre2));memset(dist,0,sizeof(dist));memset(visit,0,sizeof(visit));memset(ansetor,0,sizeof(ansetor));for(i=0;i<m;i++){scanf("%d%d%d%s",&a,&b,&c,ch);Add_edge(a,b,c);   //**如果树有结构的才用fathernum[]Add_edge(b,a,c);   //**}scanf("%d",&k);for(i=1;i<=k;i++){scanf("%d%d",&a,&b);Add_Qes(a,b,i);Add_Qes(b,a,i);}for(i=1;i<=n;i++){if(!visit[i])     //**没有访问过的点开始访问,如果树有结构的才从fathernum[]出发!LCA(i);}for(i=1;i<=k;i++){printf("%d\n",ans[i]);}}return 0;}