tarjan

父節點和子節點

  1 #include <cstdio>
  2 #include <cstdlib>
  3 #include <cmath>
  4 #include <cstring>
  5 #include <time.h>
  6 #include <string>
  7 #include <set>
  8 #include <map>
  9 #include <list>
 10 #include <stack>
 11 #include <queue>
 12
 
 #include <vector>
 13 #include <bitset>
 14 #include <ext/rope>
 15 #include <algorithm>
 16 #include <iostream>
 17 using namespace std;
 18 #define ll long long
 19 #define minv 1e-6
 20 #define inf 1e9
 21 #define pi 3.1415926536
 22 #define E  2.7182818284
 23 const ll mod=1e9+7
 
;//998244353
 24 const int maxn=1e6+10;
 25 
 26 struct node
 27 {
 28     int d,pos,c;
 29     node* next;
 30 }*e[maxn];
 31 
 32 bool vis[maxn]={0};
 33 int dfn[maxn],low[maxn],num=0,g=0,n,cond[maxn],st[maxn],fa[maxn];
 34 
 35 void dfs(int d)
 36 {
 37     int dd;
 38     node* p=e[d];
 39     dfn[d]=low[d]=++num;
 
 40     vis[d]=1;
 41     st[++g]=d;
 42     while (p)
 43     {
 44         dd=p->d;
 45         if (!vis[dd])
 46         {
 47             fa[dd]=d;
 48             dfs(dd);
 49             low[d]=min(low[d],low[dd]);
 50             cond[p->pos]=p->c;
 51         }
 52         //lack whether in stack (not necessarily)
 53         else if (fa[d]!=dd)
 54         {
 55             low[d]=min(low[d],dfn[dd]);
 56             if (dfn[d]>dfn[dd])
 57                 cond[p->pos]=p->c;
 58         }
 59         p=p->next;
 60     }
 61     if (dfn[d]==low[d])
 62     {
 63         if (g!=n || st[1]!=d)
 64         {
 65             printf("impossible");
 66             exit(0);
 67         }
 68     }
 69 }
 70 
 71 int main()
 72 {
 73     node* p;
 74     int m,i,x,y;
 75     scanf("%d%d",&n,&m);
 76     for (i=1;i<=m;i++)
 77     {
 78         scanf("%d%d",&x,&y);
 79         p=(node*) malloc (sizeof(node));
 80         p->d=y;
 81         p->pos=i;
 82         p->c=0;
 83         p->next=e[x];
 84         e[x]=p;
 85 
 86         p=(node*) malloc (sizeof(node));
 87         p->d=x;
 88         p->pos=i;
 89         p->c=1;
 90         p->next=e[y];
 91         e[y]=p;
 92     }
 93     fa[1]=0;
 94     dfs(1);
 95     for (i=1;i<=m;i++)
 96         printf("%d",cond[i]);
 97     return 0;
 98 }
 99 /*
100 3 2
101 1 2
102 2 3
103 */

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