HDU2586 How far away ？（求LCA）

本题的考点是LCA最近公共祖先

How far away ？（HDU2586）

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19934 Accepted Submission(s): 7806

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

我用的是LCA问题的离线算法的Tarjan算法求解的

#include<cstdio>  
#include<iostream>  
#include<cstring>  
#include<cmath>  
#include<algorithm>  
using namespace std;  
const int N = 40000+10;  
const int M = 220;  
int head[N],head1[N],dis[N],LCA[N],father[N];  
bool vis[N];  
int n,m,cnt;  
struct Edge{  
    int from,to;  
    int next;  
    int val;  
};  
struct Edge1{  
    int u,v;  
    int num;  
    int next;  
};  
Edge edge[N*2];  
Edge1 edge1[M*2];  
void add_edge(int u,int v,int val){  
    edge[cnt].from=u;  
    edge[cnt].to=v;  
    edge[cnt].val=val;  
    edge[cnt].next=head[u];  
    head[u]=cnt++;  
}  
void add_ans(int u,int v,int num){  
    edge1[cnt].u=u;  
    edge1[cnt].v=v;  
    edge1[cnt].num=num;  
    edge1[cnt].next=head1[u];  
    head1[u]=cnt++;  
}   
int find(int x){  
    //return x==father[x]?x:find(father[x]);   
    int r=x;  
    while(r!=father[r]){  
        r=father[r];  
    }  
    return r;  
//  if(x!=father[x]){  
//      x=father[x];  
//  }  
//  return father[x];  
}  
void tarjan(int k){  
    vis[k]=1;  
    father[k]=k;  
    for(int i=head1[k];i!=-1;i=edge1[i].next){  
        int v=edge1[i].v;  
        if(vis[v]){  
            LCA[edge1[i].num]=find(v);  
        }  
    }  
    for(int i=head[k];i!=-1;i=edge[i].next){  
        int to=edge[i].to;  
        if(!vis[to]){  
            dis[to]=dis[k]+edge[i].val;  
            tarjan(to);  
            father[to]=k;  
        }  
    }  
}  
int main(){  
    int T;  
    scanf("%d",&T);  
    while(T--){  
        scanf("%d%d",&n,&m);  
        int a,b,c;  
        cnt=0;  
        memset(head,-1,sizeof(head));  
        memset(dis,0,sizeof(dis));   
        for(int i=0;i<n-1;i++){  
            scanf("%d%d%d",&a,&b,&c);  
            add_edge(a,b,c);  
            add_edge(b,a,c);  
        }  
        dis[1]=0;  
        cnt=0;  
        memset(head1,-1,sizeof(head1));  
        for(int i=1;i<=m;i++){  
            scanf("%d%d",&a,&b);  
            add_ans(a,b,i);  
            add_ans(b,a,i);  
        }  
        memset(vis,0,sizeof(vis));  
        tarjan(1);  
        for(int i=1;i<=m*2;i+=2){  
            a=edge1[i].u;  
            b=edge1[i].v;  
            c=edge1[i].num;  
            printf("%d\n",dis[a]+dis[b]-2*dis[LCA[c]]);  
        }  
    }  
    return 0;  
}
  

 

  
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文章来源: blog.csdn.net，作者：爱玲姐姐，版权归原作者所有，如需转载，请联系作者。

原文链接：blog.csdn.net/jal517486222/article/details/79335837