LCA在线算法（hdu2586）

hdu2586

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4183    Accepted Submission(s): 1598

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

分析：

LCA最近公共祖先其实就是利用树形结构求两点间的最短路；一般题目说有n个点，且仅有n-1条边，则优先考虑LCA算法（离线或在线）

程序：

#include"stdio.h"#include"string.h"#include"stdlib.h"#define M 40009#include"string"#include"map"#include"iostream"using namespace std;struct st{    int u,v,next,w;}edge[M];int head[M],f[M],rank[M],use[M],dis[M],t;void init(){    t=0;    memset(head,-1,sizeof(head));}void add(int u,int v,int w){    edge[t].u=u;    edge[t].v=v;    edge[t].w=w;    edge[t].next=head[u];    head[u]=t++;}void dfs(int u){    int i;    use[u]=1;    for(i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(!use[v])        {            rank[v]=rank[u]+1;            dis[v]=dis[u]+edge[i].w;            dfs(v);        }    }}int LCA(int u,int v){    if(u==v)        return u;    else if(rank[u]>rank[v])        return LCA(f[u],v);    else        return LCA(u,f[v]);}int main(){    int T,i,a,b,c,m,n;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        for(i=1;i<=n;i++)            f[i]=i;        init();        for(i=1;i<n;i++)        {            scanf("%d%d%d",&a,&b,&c);            add(a,b,c);            f[b]=a;        }        memset(dis,0,sizeof(dis));        memset(rank,0,sizeof(rank));        memset(use,0,sizeof(use));        for(i=1;i<=n;i++)            if(f[i]==i)            dfs(i);        while(m--)        {            scanf("%d%d",&a,&b);            int ans=LCA(a,b);            printf("%d\n",dis[a]+dis[b]-2*dis[ans]);        }    }    return 0;}