HDU How far away ？--LCA

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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

3 2

1 2 10

3 1 15

1 2

2 3

2 2

1 2 100

1 2

2 1

Sample Output

100

Source

ECJTU 2009 Spring Contest

题目大意：

共有t组数据

给你n个点，n-1条双向边

有m次询问问x~y点之间的距离是多少

 1 #include <cstring> 2 #include <ctype.h> 3 #include <cstdio> 4  5 const int MAXN=40010; 6  7 int t,n,m; 8  9 struct node {10     int to;11     int next;12     int val;13 };14 node e[MAXN<<1];15 16 int head[MAXN],tot;17 18 int deep[MAXN],f[MAXN][21],dis[MAXN];19 20 inline void read(int&x) {21     int f=1;register char c=getchar();22     for(x=0;!isdigit(c);c=='-'&&(f=-1),c=getchar());23     for(;isdigit(c);x=x*10+c-48,c=getchar());24     x=x*f;25 }26 27 inline void add(int x,int y,int v) {28     e[++tot].to=y;29     e[tot].val=v;30     e[tot].next=head[x];31     head[x]=tot;32 }33 34 void dfs(int u) {35     deep[u]=deep[f[u][0]]+1;36     for(int i=head[u];i!=-1;i=e[i].next) {37         int to=e[i].to;38         if(!deep[to]&&to) {39             f[to][0]=u;40             dis[to]=dis[u]+e[i].val;41             dfs(to);42         }43     }44     return;45 }46 47 inline void swap(int&x,int&y) {48     int t=x;49     x=y;y=t;50     return;51 }52 53 inline int LCA(int x,int y) {54     if(deep[x]<deep[y]) swap(x,y);55     int t=deep[x]-deep[y];56     for(int i=20;i>=0;--i) 57       if(deep[f[x][i]]>=deep[y]) x=f[x][i];58     if(x==y) return x;59     for(int i=20;i;--i) 60       if(f[x][i]!=f[y][i])61         x=f[x][i],y=f[y][i];62     return f[x][0];63     64 }65 66 inline void pre() {67     tot=1;68     memset(head,-1,sizeof head);    69     memset(deep,0,sizeof deep);70     memset(dis,0,sizeof dis);71 }72 73 int hh() {74     int x,y,z;75     read(t);76     while(t--) {77         read(n);read(m);78         pre();79         for(int i=1;i<n;++i) {80             read(x);read(y);read(z);81             add(x,y,z);82             add(y,x,z);83         }84         dfs(1);85         for(int j=1;j<=20;++j) 86           for(int i=1;i<=n;++i) 87             f[i][j]=f[f[i][j-1]][j-1];88         for(int i=1;i<=m;++i) {89             read(x);read(y);90             int lca=LCA(x,y);91             printf("%d\n",dis[x]+dis[y]-2*dis[lca]);92         }93     }94     return 0;95 }96 97 int sb=hh();98 int main() {;}

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