hdu 2586(LCA+并查集)

How far away ?Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6734Accepted Submission(s): 2498

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

Source

Recommend

lcy

就是一个最近公共祖先问题,tarjan算法,,说白了只是一个dfs。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;#define maxn 40010typedef struct node{int v,d;int next;}Node;int index;int head[maxn];Node edge[2*maxn];int index1;int head1[maxn];Node edge1[2*maxn];int res[210][3];int f[maxn];int vis[maxn];int dis[maxn];int n,m;void add(int u,int v,int d){index++;edge[index].v=v;edge[index].d=d;edge[index].next=head[u];head[u]=index;}void add1(int u,int v,int d){index1++;edge1[index1].v=v;edge1[index1].d=d;edge1[index1].next=head1[u];head1[u]=index1;}int find(int x){if(x!=f[x]){f[x]=find(f[x]);return f[x];}return x;}void tarjan(int u){vis[u]=1;f[u]=u;for(int p=head1[u];p;p=edge1[p].next){if(vis[edge1[p].v]){res[edge1[p].d][2]=find(edge1[p].v);}}for(int p=head[u];p;p=edge[p].next){if(vis[edge[p].v]==0){dis[edge[p].v]=dis[u]+edge[p].d;//if(dis[edge[p].v]){// printf("%d\n",dis[edge[p].v]);//}tarjan(edge[p].v);f[edge[p].v]=u;}}}int main(){int T;scanf("%d",&T);while(T–){scanf("%d%d",&n,&m);memset(head,0,sizeof(head));index=0;for(int i=1;i<n;i++){int u,v,d;scanf("%d%d%d",&u,&v,&d);add(u,v,d);add(v,u,d);}memset(head1,0,sizeof(head1));index1=0;for(int i=0;i<m;i++){int u,v;scanf("%d%d",&u,&v);res[i][0]=u;res[i][1]=v;add1(u,v,i);add1(v,u,i);}memset(vis,0,sizeof(vis));dis[1]=0;tarjan(1);//for(int i=1;i<=n;i++){// printf("%d ",dis[i]);//}//printf("\n");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;}

你所缺少的部分,也早已被我用想像的画笔填满。

hdu 2586(LCA+并查集)

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