hdu 2586 How far away ？<vector>

How far away ？

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2113 Accepted Submission(s): 776

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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#include<stdio.h> #include<mem.h> #include<vector> using namespace std; struct node {     int oth,dis; }temp; const int maxt=40004; vector<node> tu[40004]; vector<node> qu[40004]; bool flag[maxt]; int f[maxt]; int path[maxt]; int ans[maxt]; int find(int x) {     while(x!=f[x])     {         x=f[x];     }     return x; } int LCA(int k) {     int i,j;     int oth1,oth2;     for(i=0;i<tu[k].size();i++)     {         oth1=tu[k][i].oth;         if(flag[oth1]==false)         {             flag[oth1]=true;             path[oth1]=path[k]+tu[k][i].dis;             LCA(oth1);             f[oth1]=k;             for(j=0;j<qu[oth1].size();j++)             {                 oth2=qu[oth1][j].oth;                 if(flag[oth2]==true&&ans[qu[oth1][j].dis]==0)                 {                     if(oth1!=oth2)                     {                         ans[qu[oth1][j].dis]=path[oth1]+path[oth2]-2*path[find(oth2)];                     }                 }             }         }     }     return 0; } int main() {     int t,i,m,n;     int a,b,c;     scanf("%d",&t);     while(t>=1)     {         scanf("%d%d",&n,&m);         for(i=1;i<=n;i++)         {             f[i]=i;             flag[i]=false;             path[i]=0;             ans[i]=0;             tu[i].clear();             qu[i].clear();         }         for(i=1;i<n;i++)         {             scanf("%d%d%d",&a,&b,&c);             temp.oth=a;             temp.dis=c;             tu[b].push_back(temp);             temp.oth=b;             tu[a].push_back(temp);         }         for(i=1;i<=m;i++)         {             scanf("%d%d",&a,&b);             temp.oth=a;             temp.dis=i;             qu[b].push_back(temp);             temp.oth=b;             qu[a].push_back(temp);         }         flag[1]=true;         path[1]=0;         LCA(1);         for(i=1;i<=m;i++)         {             printf("%d\n",ans[i]);         }         t--;     } }