HDU4674 Trip Advisor

Problem Description

There is a strange country somewhere which its transportation network was built following some weird rules: Consider the transportation network as a connected undirected graph G with N nodes and M edges(nodes indicate cities,and people can travel from cities to cities by roads connecting them),the condition that every node in G is in at most one simple cycle holds.
One day Q people of the country want to make a travel, i_th people is located at city U_i, wants to move to city V_i(U_i can be equal to V_i),and he especially loves city P_i,so he wonders if there is a path that:

1. starts at U_i;
2. ends at V_i;
3. for every pair of adjacent cities in path,there is a road connecting them.
4. visits every city at most once.
5. visits city P_i;

As a trip advisor,your task is to tell everybody whether there is a path satsifying all his requirements.

 

Input

The input contains several test cases, terminated by EOF. Most of test cases are rather small.
Each testcase contains M + Q + 2 lines. first line contains two integers N, M (1 <= N <= 100000, 1 <= M <= 150000), the number of cities and roads; next M lines each contains two integer b, e indicates that there is a bidirectional-road connecting city b and city e;next line an integer Q (1 ≤ Q ≤ 100000) indicating number of people in that country that want to make a travel, next Q lines each contains three integers U_i, V_i, P_i, denotes the query as we mentioned above.

 

Output

For each test case output Q lines, for i-th query, if there exists such path, output one line "Yes" without quotes, otherwise output one line "No" without quotes.

 

Sample Input

6 7
1 2
1 2
3 4
2 4
5 6
5 6
4 5
5
1 6 3
1 6 4
1 2 5
2 2 2
2 2 3

 

Sample Output

No
Yes
No
Yes
No

 

Source

2013 Multi-University Training Contest 7

 //#pragma comment(linker, "/STACK:65536000")
 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 #include <vector>
 #include <map>
 #include <math.h>
 #include <queue>
 #include <set>
 #include <algorithm>
 #define clr(a,b) memset(a,b,sizeof(a))
 #define mpr(a,b) make_pair(a,b)
 #define ll long long
 #define eps 1e-6
 using namespace std;

 const int N=,M=;

 /*
 * HDU4674
 * 一个无向图，每个点最多只属于一个简单环，询问从a到b的简单路径能不能路过c
 * 离线处理，先缩点，假设缩点完后a为A，b为B，c为C
 * 对每个询问分别求出lca(A,B),lca(A,C),lca(B,C)
 * 还要判断两个点求lca的时候，对lca的那个点，进出的联通块分别是哪个点，这个
 * 在做完一边lca后再dfs一边就好了
 * come记录从父亲节点进入这个联通块的是哪个点
 * out数组在dfs中记录当前祖先的状态
 * gra数组用来缩点建图，first保存连到下一个联通块的编号，
                second保存从该联通块该边连出去的是哪个点
 * qu数组用来lca询问
 * tlca记录每个lca的值
 * pt表示两个点连到lca的时候，进出lca那个联通块的点，-1说明就在lca上
 */
 int n,m,q,eid,id,now,top;
 int head[N],ed[M],nxt[M];
 int dfn[N],low[N],gid[N];
 int sta[N],block[N];
 vector<pair<int,int> >gra[N],qu[N];
 int f[N],vis[N],come[N],out[N],fa[N];
 int tlca[*N],pt[*N][];
 int aa[N],bb[N],cc[N];

 int findfa(int s){return s==fa[s]?s:fa[s]=findfa(fa[s]);}

 void addedge(int s,int e){
      ed[eid]=e;nxt[eid]=head[s];head[s]=eid++;
 }

 void tarjan(int s,int f,int b){
      dfn[s]=low[s]=++now;
      sta[top++]=s;block[s]=b;
      for(int i=head[s];~i;i=nxt[i]){
           int e=ed[i];
           if(i==f||i==(f^))continue;
           if(!dfn[e]){
                tarjan(e,i,b);
                low[s]=min(low[s],low[e]);
           }else
                low[s]=min(low[s],dfn[e]);
      }
      if(low[s]==dfn[s]){
           id++;
           while(top){
                int k=sta[--top];
                gid[k]=id;
                if(k==s)return ;
           }
      }
 }

 void lca(int s,int f){
      fa[s]=s;
      for(int i=;i<(int)gra[s].size();i++){
           int e=gra[s][i].first;
           if(e==f)continue;
           lca(e,s);
           fa[findfa(e)]=s;
      }
      vis[s]=;
      for(int i=;i<(int)qu[s].size();i++){
           int e=qu[s][i].first,d=qu[s][i].second;
           if(vis[e])tlca[d]=findfa(e);
      }
 }

 void dfs(int s,int f){
      for(int i=;i<(int)qu[s].size();i++){
           int d=qu[s][i].second;
           int k=(tlca[d]==s)?-:out[tlca[d]];
           if(~pt[d][])pt[d][]=k;
           else pt[d][]=k;
      }
      for(int i=;i<(int)gra[s].size();i++){
           int e=gra[s][i].first,g=gra[s][i].second;
           if(e==f){
                come[s]=g;continue;
           }
           out[s]=g;
           dfs(e,s);
      }
 }

 int main(){
 //     freopen("/home/axorb/in","r",stdin);
      while(~scanf("%d%d",&n,&m)){
           eid=;clr(head,-);
           for(int i=;i<m;i++){
                int a,b;scanf("%d%d",&a,&b);
                addedge(a,b);addedge(b,a);
           }
           id=now=top=;
           for(int i=;i<=n;i++)dfn[i]=;
           int cnt=;
           for(int i=;i<=n;i++)
                if(!dfn[i])tarjan(i,-,++cnt);
           for(int i=;i<=id;i++){
                gra[i].clear();qu[i].clear();
           }
           for(int i=;i<=n;i++)
                for(int j=head[i];~j;j=nxt[j]){
                     int s=gid[i],e=gid[ed[j]];
                     if(s!=e)gra[s].push_back(mpr(e,i));
                }
           clr(f,);
           scanf("%d",&q);
           for(int i=;i<q;i++){
                int a,b,c;scanf("%d%d%d",&a,&b,&c);
                //如果3个点不能到达，则答案肯定否定
                if(block[a]!=block[b]){f[i]=;continue;}
                if(block[b]!=block[c]){f[i]=;continue;}
                if(block[a]!=block[c]){f[i]=;continue;}
                //如果起点和终点相同，那个路过的点也必须相同
                if(a==b&&a!=c){f[i]=;continue;}
                int s=gid[a],e=gid[b],k=gid[c];
                qu[s].push_back(mpr(e,i*));
                qu[e].push_back(mpr(s,i*));
                qu[s].push_back(mpr(k,i*+));
                qu[k].push_back(mpr(s,i*+));
                qu[e].push_back(mpr(k,i*+));
                qu[k].push_back(mpr(e,i*+));
                aa[i]=a;bb[i]=b;cc[i]=c;
           }
           for(int i=;i<=id;i++)vis[i]=;
           for(int i=;i<q*;i++)
                pt[i][]=pt[i][]=-;
           for(int i=;i<=id;i++)
                if(!vis[i]){
                     lca(i,-);
                     dfs(i,-);
                }
           for(int i=;i<q;i++){
                if(f[i]){
                     puts("No");continue;
                }
                if(gid[aa[i]]==gid[bb[i]]){//如果A=B，那个C必须等于A
                     if(gid[cc[i]]==gid[aa[i]])puts("Yes");
                     else puts("No");
                }
                else if(gid[cc[i]]==tlca[i*]){//如果C是lca(A,B)
                     if(pt[i*][]==-){
                          //如果AB中有一个等于C，那么当那个点是出口且C不是出口的时候，C就不能到达
                          int k=(gid[aa[i]]==gid[cc[i]])?aa[i]:bb[i];
                          if(k==pt[i*][]&&cc[i]!=k) puts("No");
                          else puts("Yes");
                     }else{
                          //如果AB都不是C，那个当AB进出C的是同一个点且不是C，C就不能到达
                          if(pt[i*][]==pt[i*][]&&cc[i]!=pt[i*][]) puts("No");
                          else puts("Yes");
                     }
                }else if(tlca[i*+]==gid[cc[i]]&&tlca[i*+]==tlca[i*]){
                     //如果C是A的祖先，且lca(A,B)=lca(C,B)
                     //那么当父亲到C的点和C连出去到A的点是同一个点且不是C，那个C就不能到达
                     int k=(pt[i*+][]==-)?aa[i]:pt[i*+][];
                     if(come[gid[cc[i]]]==k&&k!=cc[i])puts("No");
                     else puts("Yes");
                }else if(tlca[i*+]==gid[cc[i]]&&tlca[i*+]==tlca[i*]){
                     //同上
                     int k=(pt[i*+][]==-)?bb[i]:pt[i*+][];
                     if(come[gid[cc[i]]]==k&&k!=cc[i])puts("No");
                     else puts("Yes");
                }else puts("No");
           }
      }
 }