How far away ？--LCA

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13746    Accepted Submission(s): 5158

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

23 21 2 103 1 151 22 32 21 2 1001 22 1

 

Sample Output

1025100100

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2586

最近刚学LCA，对于Tarjan的离线算法掌握的自认为差不多，然后去做了这个题，然后发现，不会做。。。然后去看了题解，一看就懂了，然后自己敲了一遍，最后又忘记*2了，脑子智障了，找了将近半小时，一个小时多点敲出来的这个题，有点慢了，不能自傲啊，果然一自傲和自满我就死了，要注意了。

题意大概是说有n个点，m次询问，每次询问让你求u,v之间的最小距离。

我们设res[][]数组记录询问，res[i][0]代表第i次询问起点u，res[i][1]代表第i次询问终点v，res[i][2]代表u,v的最近公共祖先，我们再开一个dis[]数组，代表此节点到根节点的距离，这样u,v的最近距离就是dis[res[i][0]]+dis[res[i][1]]-2*dis[res[i][2]];答案就出来了

LCA的求法具体见我上一篇博客

代码：

#include <cstdio>#include <cstring>#include <iostream>using namespace std;const int N=500000;struct node{    int w,v;    struct node *pre;}edge1[N],edge2[N],*p1[N],*p2[N];int nEdge1,nEdge2;int res[N][5];bool vis[N];int pre[N];int dis[N];void connect(int u,int v,int w,node edge[],node *p[],int &nEdge){    nEdge++;    edge[nEdge].v=v;    edge[nEdge].w=w;    edge[nEdge].pre=p[u];    p[u]=edge+nEdge++;    nEdge++;    edge[nEdge].v=u;    edge[nEdge].w=w;    edge[nEdge].pre=p[v];    p[v]=edge+nEdge++;}void Init(int n){    memset(p1,0,sizeof(p1));    memset(p2,0,sizeof(p2));    memset(vis,false,sizeof(vis));    memset(dis,0,sizeof(dis));    memset(res,0,sizeof(res));    nEdge1=0;    nEdge2=0;}int Find(int x){    if(x!=pre[x])        return pre[x]=Find(pre[x]);    return x;}void Tarjan(int x){    vis[x]=true;    pre[x]=x;    for(node *p=p2[x];p;p=p->pre){        if(vis[p->v]){            res[p->w][2]=Find(p->v);        }    }    for(node *p=p1[x];p;p=p->pre){        if(!vis[p->v]){            dis[p->v]=dis[x]+p->w;            Tarjan(p->v);            pre[p->v]=x;        }    }}int main(){    int t;    scanf("%d",&t);    while(t--){        int n,m;        scanf("%d%d",&n,&m);        Init(n);        for(int i=0;i<n-1;i++){            int u,v,w;            scanf("%d%d%d",&u,&v,&w);            connect(u,v,w,edge1,p1,nEdge1);        }        for(int i=0;i<m;i++){            int u,v;            scanf("%d%d",&u,&v);            res[i][0]=u;            res[i][1]=v;            connect(u,v,i,edge2,p2,nEdge2);        }        Tarjan(1);        for(int i=0;i<m;i++){            printf("%d\n",dis[res[i][0]]+dis[res[i][1]]-2*dis[res[i][2]]);        }    }    return 0;}