How far away ？
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10312 Accepted Submission(s): 3743
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.
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
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
25
100
100
#include <bits/stdc++.h>
using namespace std ;
const int N = ; typedef pair<int,int> pii ;
#define X first
#define Y second
int n , m ;
int eh[N] , nxt[N<<] , et[N<<] , ew[N<<] , tot ;
int fa[N] , vis[N] , anc[N] ;
int ux[N] , vx[N] , wx[N] ; vector<pii>qry[N] ; void init() { for( int i = ; i <= n ; ++i ) {
qry[i].clear() ;
vis[i] = ;
} tot = ;
memset( eh , - , sizeof eh ) ; } void addedge( int u , int v , int w ) {
et[tot] = v , nxt[tot] = eh[u] , ew[tot] = w , eh[u] = tot++ ;
} int find(int x){ return fa[x] = ( fa[x]==x?x:find(fa[x]));}
void un( int x , int y ) {
x = find(x);
y = find(y);
if(x==y)return ;
fa[y]=x;
}
void dfs1( int u , int pre ) { vis[u] = ;
fa[u] = u ;
for( int i = eh[u] ; ~i ; i = nxt[i] ) {
int v = et[i] ; if( v == pre ) continue ;
dfs1( v , u ) ;
un( u , v );
} int siz = qry[u].size() ;
for( int i = ; i < siz ; ++i ) {
if( vis[qry[u][i].X] ) {
anc[qry[u][i].Y] = find( qry[u][i].X );
}
}
} int dis[N] ; void dfs2( int u , int pre , int d ) {
dis[u] = d ;
for( int i = eh[u] ; ~i ; i = nxt[i] ) {
int v = et[i] ; if( v == pre ) continue ;
dfs2( v , u , d + ew[i] ) ;
}
} int main() {
#ifdef LOCAL
freopen("in.txt","r",stdin);
#endif // LOCAL
int _ ; scanf("%d",&_) ;
while( _-- ) { scanf("%d%d",&n,&m) ;
init() ;
for( int i = ; i < n ; ++i ) {
int u , v , w ;
scanf("%d%d%d",&u,&v,&w) ;
addedge( u , v , w ) ;
addedge( v , u , w ) ;
} for( int i = ; i < m ; ++i ) {
scanf("%d%d",&ux[i],&vx[i]) ;
qry[ ux[i] ].push_back( pii( vx[i] , i ) ) ;
qry[ vx[i] ].push_back( pii( ux[i] , i ) ) ;
} dfs1( , ) ;
dfs2( , , ) ; for( int i = ; i < m ; ++i ) {
printf("%d\n",dis[ux[i]]+dis[vx[i]]-*dis[anc[i]]);
}
}
return ;
}