【杭电oj2586】How far away ？

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13230    Accepted Submission(s): 4956

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

ECJTU 2009 Spring Contest

 

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tarjan离线算法解决LCA问题，dfs+邻接表+并查集。核心代码是dfs部分，对dfs理解深刻的话应该能看懂，实在不懂就自己心算一下详细过程。

其他解决LCA问题的方法还有很多，比如转化为RMQ问题，看看自己对那个算法理解深刻去应用。

#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;}