Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3423 Accepted Submission(s): 1274
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
Recommend lcy
#include<stdio.h>
#include<string.h>
#include<vector>
#define GS 40025
using namespace std;
vector<int> G[GS],E[GS];
int father[GS],Len[GS],Dep[GS];
int N;
void build(int R)
{
for (int i=0;i<G[R].size();i++)
{
int v=G[R][i];
if (v==father[R]) continue;
father[v]=R;
build(v);
}
}
void dfs(int R)
{
for (int i=0;i<G[R].size();i++)
{
int v=G[R][i];
if (v==father[R]) continue;
Dep[v]=Dep[R]+1;
Len[v]=Len[R]+E[R][i];
dfs(v);
}
}
int LCA(int a,int b)
{
if (a==b) return a;
if (father[a]==b) return b;
else if (father[b]==a) return a;
if (Dep[a]<Dep[b]) return LCA(a,father[b]);
else return LCA(father[a],b);
}
int main()
{
int T,M;
scanf("%d",&T);
while (T--)
{
scanf("%d%d",&N,&M);
for (int i=1;i<=N;i++) G[i].clear();
for (int i=1;i<=N;i++) E[i].clear();
for (int i=1;i<N;i++)
{
int x,y,c;
scanf("%d%d%d",&x,&y,&c);
G[x].push_back(y);
G[y].push_back(x);
E[x].push_back(c);
E[y].push_back(c);
}
memset(father,-1,sizeof(father));
build(1);
Dep[1]=0;
Len[1]=0;
dfs(1);
while (M--)
{
int x,y;
scanf("%d%d",&x,&y);
int F=LCA(x,y);
printf("%d\n",Len[x]+Len[y]-2*Len[F]);
}
}
return 0;
}