Poj1986 distance queries (LCA)

Portal:

Http://poj.org/problem? Id = 1986

Distance queries

Time limit:2000 ms		Memory limit:30000 K
Case time limit:1000 ms

Description

Farmer John's cows refused to run in his marathon since he chose a path much too long for their leisurely lifestyle. he therefore wants to find a path of a more reasonable length. the input to this problem consists of the same input as in "navigation nightmare", followed by a line containing a single integer k, followed by K "distance queries ". each distance query is a line of input containing two I Ntegers, giving the numbers of two farms between which FJ is interested in computing distance (measured in the length of the roads along the path between the two farms ). please answer FJ's distance queries as quickly as possible!

Input

* Lines 1 .. 1 + M: same format as "navigation nightmare"

* Line 2 + M: A single integer, K. 1 <= k <= 10,000

* Lines 3 + M .. 2 + M + K: each line corresponds to a distance query and contains the indices of two farms.

Output

* Lines 1. K: For each distance query, output on a single line an integer giving the appropriate distance.

Sample Input

7 61 6 13 E6 3 9 E3 5 7 S4 1 3 N2 4 20 W4 7 2 S31 61 42 6

Sample output

13336

Hint

Farms 2 and 6 are 20 + 3 + 13 = 36 apart.

Source

Usaco 2004 February

 1 #include<set> 2 #include<queue> 3 #include<cstdio> 4 #include<cstdlib> 5 #include<cstring> 6 #include<iostream> 7 #include<algorithm> 8 using namespace std; 9 const int N = 100010;10 #define For(i,n) for(int i=1;i<=n;i++)11 #define Rep(i,l,r) for(int i=l;i<=r;i++)12 13 struct Edge{14     int t,next,w;15 }E[N*2];16 vector< pair<int,int> > Q[N];17 int s,t,w,Es,ans[N],head[N],fa[N],d[N];18 bool vis[N];19 20 int find(int i){21     if(fa[i]==i) return i;22     else         return find(fa[i]);23 }24 25 void makelist(int s,int t,int w){26     E[Es].t = t; E[Es].w = w;E[Es].next = head[s];27     head[s] = Es++;28 }29 30 void LCA(int i,int w){31     d[i] = w;fa[i] = i;32     vis[i] = true;33     for(int p=head[i];p!=-1;p=E[p].next)34         if(!vis[E[p].t]){35             LCA(E[p].t,w+E[p].w);36             fa[E[p].t] = i;37         }38     int  len = Q[i].size();39     for(int j=0;j<len;j++){40         if(vis[Q[i][j].first])41             ans[Q[i][j].second] = w + d[Q[i][j].first] - 2 * d[find(Q[i][j].first)]; 42     }43 }44 int n,m,qn;45 int main(){46     memset(head,-1,sizeof(head));47     scanf("%d%d",&n,&m);48     For(i,m){49         scanf("%d%d%d %*c",&s,&t,&w);50         makelist(s,t,w);makelist(t,s,w);51     }52     scanf("%d",&qn);53     For(i,qn){54         scanf("%d%d",&s,&t);55         Q[s].push_back(make_pair(t,i));Q[t].push_back(make_pair(s,i));56     }57     LCA(1,0);58     For(i,qn) printf("%d\n",ans[i]);59     return 0;60 }