Traffic Real Time Query System 圆方树+LCA

题目描述

City C is really a nightmare of all drivers for its traffic jams. To solve the traffic problem, the mayor plans to build a RTQS (Real Time Query System) to monitor all traffic situations. City C is made up of N crossings and M roads, and each road connects two crossings. All roads are bidirectional. One of the important tasks of RTQS is to answer some queries about route-choice problem. Specifically, the task is to find the crossings which a driver MUST pass when he is driving from one given road to another given road.

Input

There are multiple test cases.

For each test case:

The first line contains two integers N and M, representing the number of the crossings and roads.

The next M lines describe the roads. In those M lines, the i th line (i starts from 1)contains two integers X i and Y i, representing that road i connects crossing X i and Y i (X i≠Y i).

The following line contains a single integer Q, representing the number of RTQs.

Then Q lines follows, each describing a RTQ by two integers S and T(S≠T) meaning that a driver is now driving on the roads and he wants to reach roadt . It will be always at least one way from roads to roadt.

The input ends with a line of “0 0”.

Please note that: 0<N<=10000, 0<M<=100000, 0<Q<=10000, 0<X i,Y i<=N, 0<S,T<=M

Output

For each RTQ prints a line containing a single integer representing the number of crossings which the driver MUST pass.

样例

Sample Input

Sample Output

0

1

分析

题意：一个城市有n个路口，m条无向公路。现在要求从第S条路到第T条路必须经过的点有几个。

这道题和压力那一道题比较像，但是比那一道题要简单一些。

题目中要求从一条路到另一条路的必经点，我们先不去考虑边，我们先只去考虑点

如果是点的话那就很好说了，我们用Tarjan对点双进行缩点，使之成为一棵圆方树

缩点之后的树一定是圆点与方点交替分布的，我们算出两点之间经过的圆点个数就可以了

要求经过的圆点个数，我们要先求两点之间的距离

最后再把这个距离除以2，再向下取整减去1就是我们想要的结果

\(ans(xx,yy)=(dep[xx]+dep[yy]-dep[LCA(xx,yy)]\times2)/2-1\)

这个结果是怎么来的呢，我们简单地推导一下

很显然的是，原先的的边所连接的两个点在新的圆方树中都是圆点

而且这两个点的的最近公共祖先不是方点就是圆点

我们设这两个点分别为xx、yy

1、如果它们的最近公共祖先是圆点

xx和它的最近公共祖先间圆点的个数=\((dep[xx]-dep[LCA(xx,yy)])/2\)

yy和它的最近公共祖先间圆点的个数=\((dep[yy]-dep[LCA(xx,yy)])/2\)

因为它们的最近公共祖先作为圆点被计算了两次，所以两式相加后还要减去1

\(ans（xx,yy）=(dep[xx]-dep[LCA(xx,yy)])/2+(dep[yy]-dep[LCA(xx,yy)])/2-1\)

化简得到\(ans(xx,yy)=(dep[xx]+dep[yy]-dep[LCA(xx,yy)]\times2)/2-1\)

2、如果它们的最近公共祖先是方点

xx和它的最近公共祖先间圆点的个数=\((dep[xx]-dep[LCA(xx,yy)]-1)/2\)

yy和它的最近公共祖先间圆点的个数=\((dep[yy]-dep[LCA(xx,yy)]- 1)/2\)

两式相加

\(ans（xx,yy）=(dep[xx]-dep[LCA(xx,yy)]-1)/2+(dep[yy]-dep[LCA(xx,yy)]-1)/2\)

化简得到\(ans(xx,yy)=(dep[xx]+dep[yy]-dep[LCA(xx,yy)]\times2)/2-1\)

因此最终的结果为\(ans(xx,yy)=(dep[xx]+dep[yy]-dep[LCA(xx,yy)]\times2)/2-1\)

但是我们要求的是边到边之间经过的圆点的个数，这怎么办呢

其实我们只要把每条边的两个顶点分别枚举求最大值就可以了

代码

#include<cstdio>

#include<iostream>

#include<cstring>

#include<algorithm>

#include<vector>

using namespace std;

const int maxd=40005,maxb=400005;

struct asd{

    int from,to,next;

}b[maxb],b2[maxb];

int tot=1,t2=1,head[maxd],h2[maxd],nn;

void ad(int aa,int bb){

    b[tot].from=aa;

    b[tot].to=bb;

    b[tot].next=head[aa];

    head[aa]=tot++;

}

void ad2(int aa,int bb){

    b2[t2].from=aa;

    b2[t2].to=bb;

    b2[t2].next=h2[aa];

    h2[aa]=t2++;

}

int dfn[maxd],low[maxd],dfnc;

int n,m,sta[maxd],top;

void tarjan(int now,int fa){

    dfn[now]=low[now]=++dfnc;

    sta[++top]=now;

    for(int i=head[now];i!=-1;i=b[i].next){

        int u=b[i].to;

        if(!dfn[u]){

            tarjan(u,now);

            low[now]=min(low[now],low[u]);

            if(dfn[now]<=low[u]){

                nn++;

                ad2(nn,now),ad2(now,nn);

                while(1){

                    int xx=sta[top--];

                    ad2(xx,nn),ad2(nn,xx);

                    if(xx==u) break;

                }

            }

        } else if(u!=fa){

            low[now]=min(low[now],dfn[u]);

        }

    }

}

int f[maxd][22],dep[maxd];

bool vis[maxd];

void dfs(int now,int fa){

    vis[now]=1;

            dep[now]=dep[fa]+1;

            f[now][0]=fa;

    for(int i=1;(1<<i)<=dep[now];i++){

        f[now][i]=f[f[now][i-1]][i-1];

    }

    for(int i=h2[now];i!=-1;i=b2[i].next){

        int u=b2[i].to;

        if(u!=fa && !vis[u]){

            dfs(u,now);

        }

    }

}

int LCA(int xx,int yy){

    if(dep[xx]>dep[yy]) swap(xx,yy);

    int len=dep[yy]-dep[xx],k=0;

    while(len){

        if(len&1){

            yy=f[yy][k];

        }

        k++,len>>=1;

    }

    if(xx==yy) return xx;

    for(int i=20;i>=0;i--){

        if(f[xx][i]==f[yy][i]) continue;

        xx=f[xx][i],yy=f[yy][i];

    }

    return f[xx][0];

}

int solve(int xx,int yy){

    return (dep[xx]+dep[yy]-2*dep[LCA(xx,yy)])/2-1;

}

int main(){

    while(scanf("%d%d",&n,&m)!=EOF && (n|m)){

        memset(head,-1,sizeof(head));

        memset(h2,-1,sizeof(h2));

        memset(&b,0,sizeof(struct asd));

        memset(&b2,0,sizeof(struct asd));

        memset(dfn,0,sizeof(dfn));

        memset(low,0,sizeof(low));

        memset(f,0,sizeof(f));

        memset(dep,0,sizeof(dep));

        memset(sta,0,sizeof(sta));

        memset(vis,0,sizeof(vis));

        tot=1,t2=1,dfnc=0,top=0,nn=n;

        for(int i=1;i<=m;i++){

            int aa,bb;

            scanf("%d%d",&aa,&bb);

            ad(aa,bb);

            ad(bb,aa);

        }

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

            if(!dfn[i]) tarjan(i,-1);

        }

        for(int i=1;i<=nn;i++){

            if(!vis[i]) dfs(i,0);

        }

        int q;

        scanf("%d",&q);

        while(q--){

            int aa,bb;

            scanf("%d%d",&aa,&bb);

            int ans1=solve(b[aa*2].from,b[bb*2].from);

            int ans2=solve(b[aa*2].to,b[bb*2].from);

            int ans3=solve(b[aa*2].from,b[bb*2].to);

            int ans4=solve(b[aa*2].to,b[bb*2].to);

            int ans=max(max(ans1,ans2),max(ans3,ans4));

            printf("%d\n",ans);

        }

    }

    return 0;

}