HDU 2586 （Tarjan)

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12442    Accepted Submission(s): 4587

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
#include <iostream>#include <cstring>#include <cstdio>using namespace std;const int N = 800010;typedef long long LL;struct node1{    int to, wight, next;}p1[N];struct node2{    int to, index, next;}p2[N];int head1[N], head2[N], visit[N], par[N];LL dist[N], res[N];int cnt1, cnt2;void init();int n, m;void add1(int u,int v,int c);void add2(int u,int v,int c);void tarjan(int u);int ser(int x);int main(){    int t;    scanf("%d", &t);    while(t--)    {        init();        scanf("%d %d", &n, &m);        init();        int x, y, cost;        for(int i=1;i<=n-1;i++)        {            scanf("%d %d %d", &x, &y, &cost);            add1(x,y,cost);            add1(y,x,cost);        }        for(int i=1;i<=m;i++)        {            scanf("%d %d", &x, &y);            add2(x,y,i);            add2(y,x,i);        }        dist[1]=0;        tarjan(1);        for(int i=1;i<=m;i++)        {            printf("%I64d\n",res[i]);        }    }    return 0;}void init(){    memset(head1,-1,sizeof(head1));    memset(head2,-1,sizeof(head2));    memset(visit, 0, sizeof(visit));    for(int i=1;i<=n;i++)    {        par[i]=i;    }    cnt1=0,cnt2=0;    return ;}void add1(int u,int v,int c){    p1[cnt1].to=v;    p1[cnt1].wight=c;    p1[cnt1].next=head1[u];    head1[u]=cnt1++;    return ;}void add2(int u,int v,int c){    p2[cnt2].to=v;    p2[cnt2].index=c;    p2[cnt2].next=head2[u];    head2[u]=cnt2++;    return ;}void tarjan(int u){    visit[u]=1;    for(int i=head1[u];i!=-1;i=p1[i].next)    {        int v=p1[i].to;        if(!visit[v])        {            dist[v]=dist[u]+p1[i].wight;            tarjan(v);            par[v]=u;        }    }    for(int i=head2[u];i!=-1;i=p2[i].next)    {        int v=p2[i].to;        if(visit[v])        {            res[p2[i].index]=dist[u]+dist[v]-2*dist[ser(v)];        }    }    return ;}int ser(int x){    int l=x, r=x, j;    while(l!=par[l]) l=par[l];    while(r!=l) j=par[r], par[r]=l, r=j;    return l;}