51nod 1076 2条不相交的路径（边双连通分量）

  1 #include<iostream>
  2 #include<algorithm>
  3 #include<cstring>
  4 #include<cstdio>
  5 #include<sstream>
  6 #include<vector>
  7 #include<stack>
  8 #include<queue>
  9 #include<cmath>
 10 #include<map>
 11 #include<set>
 12 using namespace std;
 13 typedef long long ll;
 14 const int INF = 0x3f3f3f3f;
 15 const int maxn=50000+5;
 16 const int mod=1e9+7;
 17 
 18 int n, m;
 19 int tot;
 20 int head[maxn];
 21 int ebbc[maxn],pre[maxn],eccno[maxn],low[maxn];
 22 int ebbc_cnt,dfs_clock;
 23 
 24 stack<int> S;
 25 
 26 struct node
 27 {
 28     int v,next;
 29 }e[2*maxn];
 30 
 31 void addEdge(int u, int v)
 32 {
 33     e[tot].v=v;
 34     e[tot].next=head[u];
 35     head[u]=tot++;
 36 }
 37 
 38 int tarjan(int u, int fa)
 39 {
 40     int lowu=pre[u]=++dfs_clock;
 41     S.push(u);
 42     for(int i=head[u];i!=-1;i=e[i].next)
 43     {
 44         int v=e[i].v;
 45         if(!pre[v])
 46         {
 47             int lowv=tarjan(v,u);
 48             lowu=min(lowu,lowv);
 49         }
 50         else if(v!=fa)
 51             lowu=min(lowu,pre[v]);
 52     }
 53     if(pre[u]==lowu)
 54     {
 55         ebbc_cnt++;
 56         for(;;)
 57         {
 58             int tmp=S.top(); S.pop();
 59             eccno[tmp]=ebbc_cnt;
 60             if(tmp==u)  break;
 61         }
 62     }
 63     return low[u]=lowu;
 64 }
 65 
 66 void find_ebbc()
 67 {
 68     ebbc_cnt=dfs_clock=0;
 69     memset(pre,0,sizeof(pre));
 70     for(int i=1;i<=n;i++)
 71     {
 72         if(!pre[i])  tarjan(i,-1);
 73     }
 74 }
 75 
 76 int main()
 77 {
 78     //freopen("in.txt","r",stdin);
 79     while(~scanf("%d%d",&n,&m))
 80     {
 81         tot=0;
 82         memset(head,-1,sizeof(head));
 83         for(int i=1;i<=m;i++)
 84         {
 85             int u,v;
 86             scanf("%d%d",&u,&v);
 87             addEdge(u,v);
 88             addEdge(v,u);
 89         }
 90         find_ebbc();
 91         int q;
 92         scanf("%d",&q);
 93         while(q--)
 94         {
 95             int u,v;
 96             scanf("%d%d",&u,&v);
 97             if(eccno[u]==eccno[v])  puts("Yes");
 98             else puts("No");
 99         }
100     }
101     return 0;
102 }