How far away ？
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9359 Accepted Submission(s): 3285
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.
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
题目大意：给你一棵树，边的长度，求任意两点间的最短距离。
解题思路：直接套LCA模板就行了。LCA练手。
在线：
vset[]数组含义：dfs过程中记录经过的节点编号，其实下标可以看做是时间
dep[]数组含义：表示节点的在树中的深度
first[]数组含义：dfs过程中第一次到达节点的时间
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5;
struct AdjEdge{
int to,w,next;
}adjedges[maxn];
int head[maxn];
int dis[maxn],vset[maxn],dep[maxn],d[maxn][30],first[maxn];
int tot,nn;
void init(){
tot=0;
nn=0;
memset(dep,0,sizeof(dep));
memset(head,-1,sizeof(head));
memset(dis,0,sizeof(dis));
memset(d,0,sizeof(d));
memset(first,0,sizeof(first));
}
void addedge(int _u,int _v,int _w){ //
adjedges[tot].to=_v;
adjedges[tot].w=_w;
adjedges[tot].next=head[_u];
head[_u]=tot++;
adjedges[tot].to=_u;
adjedges[tot].w=_w;
adjedges[tot].next=head[_v];
head[_v]=tot++;
}
void dfs(int _u,int _fa,int _dep){
// printf("%d %d
",_u,_dep);
dep[_u]=_dep;
vset[++nn]=_u;
first[_u]=nn;
for(int i=head[_u];i!=-1;i=adjedges[i].next){
AdjEdge & e = adjedges[i];
if(e.to!=_fa){
dis[e.to]=dis[_u]+e.w;
dfs(e.to,_u,_dep+1);
vset[++nn]=_u;
}
}
}
void ST(){
for(int i=1;i<=nn;i++)
d[i][0]=vset[i];
for(int j=1;(1<<j)<=nn;j++){
for(int i=1; i+(1<<j)-1<=nn ; i++){
if(dep[d[i][j-1]]<dep[d[i+(1<<(j-1))][j-1]])
d[i][j]=d[i][j-1];
else d[i][j]=d[i+(1 << (j-1))][j-1];
}
}
}
int RMQ(int L,int R){
int k=0;
while((1<<(k+1))<=R-L+1) k++;
if(dep[d[L][k]]<=dep[d[R-(1<<k)+1][k]])
return d[L][k];
return d[R-(1<<k)+1][k];
}
int main(){
int T,n,q;
scanf("%d",&T);
while(T--){
init();
int a,b,c;
scanf("%d%d",&n,&q);
for(int i=1;i<=n-1;i++){
scanf("%d%d%d",&a,&b,&c);
addedge(a,b,c);
}
dfs(1,-1,1);
ST();
for(int i=0;i<q;i++){
scanf("%d%d",&a,&b);
if(first[a]<=first[b]){
int tmp1=RMQ(first[a],first[b]);
printf("%d
",dis[a]+dis[b]-2*dis[tmp1]);
}else{
int tmp1=RMQ(first[b],first[a]);
printf("%d
",dis[a]+dis[b]-2*dis[tmp1]);
}
}
}
return 0;
}