HDU 2586 How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18652    Accepted Submission(s):
7268

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

2
3 2
1 2 10
3 1 15
1 2
2 3

2 2
1 2 100
1 2
2 1

 

Sample Output

10
25
100
100

 

Source

ECJTU
2009 Spring Contest

 

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带权的LCA问题

我们用g[i]表示i号节点走到根的权值

那么两个点之间的路径权值为,$g[x]+g[y]-2*g[LCA(x,y)]$

大概是这个样子

被圆圈出来的是需要减去的

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int MAXN=1e5+;
 inline int read()
 {
     char c=getchar();int x=,f=;
     while(c<''||c>'')    {if(c=='-')    f=-;c=getchar();}
     while(c>=''&&c<='')    x=x*+c-,c=getchar();return x*f;
 }
 int n,m,S=;
 int f[MAXN][],deep[MAXN],g[MAXN];
 struct node
 {
     int u,v,w,nxt;
 }edge[MAXN];
 int head[MAXN];
 int num=;
 inline void add_edge(int x,int y,int z)
 {
     edge[num].u=x;
     edge[num].v=y;
     edge[num].w=z;
     edge[num].nxt=head[x];
     head[x]=num++;
 }
 void dfs(int now)
 {
     for(int i=head[now];i!=-;i=edge[i].nxt)
         if(deep[edge[i].v]==)
         {
             deep[edge[i].v]=deep[now]+;
             f[edge[i].v][]=now;
             g[edge[i].v]=g[now]+edge[i].w;
             dfs(edge[i].v);
         }

 }
 inline void pre()
 {
     for(int i=;i<=;i++)
         for(int j=;j<=n;j++)
             f[j][i]=f[f[j][i-]][i-];
 }
 inline int LCA(int x,int y)
 {
     if(deep[x]<deep[y])    swap(x,y);
     for(int i=;i>=;i--)
         if(deep[f[x][i]]>=deep[y])
             x=f[x][i];
     if(x==y)    return x;

     for(int i=;i>=;i--)
         if(f[x][i]!=f[y][i])
             x=f[x][i],y=f[y][i];
     return f[x][];
 }
 int main()
 {
     int T=read();
     while(T--)
     {
         n=read();m=read();
         memset(head,-,sizeof(head));num=;
         memset(f,,sizeof(f));
         memset(deep,,sizeof(deep));
         for(int i=;i<=n-;i++)
         {
             int x=read(),y=read(),z=read();
             add_edge(x,y,z);
             add_edge(y,x,z);
         }
         deep[S]=;
         dfs(S);pre();
         while(m--)
         {
             int x=read(),y=read();
             printf("%d\n",g[x]+g[y]-*g[LCA(x,y)]);
         }
     }

     return ;
 }