Distance Queries

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 12846		Accepted: 4552
Case Time Limit: 1000MS

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 integers, 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!

* 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.

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

7 6

1 6 13 E

6 3 9 E

3 5 7 S

4 1 3 N

2 4 20 W

4 7 2 S

3

1 6

1 4

2 6

13

3

36

Farms 2 and 6 are 20+3+13=36 apart. 

#include <stdio.h>

#include <iostream>

#include <algorithm>

#include <cstdlib>

#include <sstream>

#include <cstring>

#include <bitset>

#include <string>

#include <deque>

#include <stack>

#include <cmath>

#include <queue>

#include <set>

#include <map>

using namespace std;

#define INF 0x3f3f3f3f

#define CLR(arr,val) memset(arr,val,sizeof(arr))

#define LC(x) (x<<1)

#define RC(x) ((x<<1)+1)

#define MID(x,y) ((x+y)>>1)

typedef pair<int,int> pii;

typedef long long LL;

const double PI=acos(-1.0);

const int N=40010;

struct edge

{

    int to;

    int nxt;

    int w;

};

struct query

{

    int to;

    int nxt;

    int id;

};

edge E[N<<1];

query Q[N<<1];

int head[N],rhead[N],tot,rtot;

int d[N],dx[N],vis[N],in[N];

int pre[N],ances[N];

void init()

{

    CLR(head,-1);

    CLR(rhead,-1);

    tot=rtot=0;

    CLR(d,0);

    for (int i=0; i<N; ++i)

    {

        pre[i]=i;

        ances[i]=0;

    }

    CLR(vis,0);

    CLR(in,0);

    CLR(dx,0);

}

int Find(int n)

{

    if(pre[n]==n)

        return n;

    return pre[n]=Find(pre[n]);

}

inline void add(int s,int t,int d)

{

    E[tot].to=t;

    E[tot].w=d;

    E[tot].nxt=head[s];

    head[s]=tot++;

}

inline void addquery(int s,int t,int id)

{

    Q[rtot].id=id;

    Q[rtot].to=t;

    Q[rtot].nxt=rhead[s];

    rhead[s]=rtot++;

}

void LCA(int u)

{

    vis[u]=1;

    ances[u]=u;

    int i,v;

    for (i=head[u]; ~i; i = E[i].nxt)

    {

        v = E[i].to;

        if(!vis[v])

        {

            LCA(v);

            pre[v]=u;

            ances[Find(u)]=u;

        }

    }

    for (i=rhead[u]; ~i; i = Q[i].nxt)

    {

        v=Q[i].to;

        if(vis[v])

            dx[Q[i].id]=d[u]+d[v]-(d[ances[Find(v)]]<<1);

    }

}

void dfs(int u,int fa,int sum)

{

    d[u]=sum;

    for (int i=head[u]; ~i; i = E[i].nxt)

    {

        int v=E[i].to;

        if(v!=fa)

            dfs(v,u,sum+E[i].w);

    }

}

int main(void)

{

    int n,m,a,b,c,i,k;

    char nouse[5];

    while (~scanf("%d%d",&n,&m))

    {

        init();

        for (i=0; i<m; ++i)

        {

            scanf("%d%d%d%s",&a,&b,&c,nouse);

            add(a,b,c);

            add(b,a,c);

            ++in[b];

        }

        scanf("%d",&k);

        for (i=0; i<k; ++i)

        {

            scanf("%d%d",&a,&b);

            addquery(a,b,i);

            addquery(b,a,i);

        }

        for (i=1; i<=n; ++i)

        {

            if(!in[i])

            {

                dfs(i,-1,0);

                LCA(i);

                break;

            }

        }

        for (i=0; i<k; ++i)

            printf("%d\n",dx[i]);

    }

    return 0;

}