POJ 1236 Network of Schools

强连通分量求入度为零和出度为零的个数；

  1 #include<iostream>
  2 #include<algorithm>
  3 #include<stdio.h>
  4 #include<stdlib.h>
  5 #include<string.h>
  6 #define maxn 110
  7 using namespace std;
  8 int head1[maxn],head2[maxn];
  9 int out[maxn],in[maxn];
 10 int mark1[maxn],mark2[maxn];
 11 int Et[maxn],vis[maxn],ss[maxn];
 12 int cont1,cont2,p,q,n,ans,t,k,num;
 13 struct node
 14 {
 15     int to,next;
 16 } E1[maxn*maxn],E2[maxn*maxn],e[maxn*maxn];
 17 void Init()
 18 {
 19     memset(head1,-1,sizeof(head1));
 20     memset(head2,-1,sizeof(head2));
 21     memset(out,0,sizeof(out));
 22     memset(in,0,sizeof(in));
 23     memset(mark1,0,sizeof(mark1));
 24     memset(mark2,0,sizeof(mark2));
 25     memset(Et,0,sizeof(Et));
 26     memset(vis,0,sizeof(vis));
 27     memset(ss,0,sizeof(ss));
 28     p=q=cont1=cont2=k=t=1;
 29 };
 30 void add(int a,int b)
 31 {
 32     E1[p].to=b,E1[p].next=head1[a],head1[a]=p++;
 33     E2[q].to=a,E2[q].next=head2[b],head2[b]=q++;
 34 }
 35 void DFS1(int u)
 36 {
 37     mark1[u]=1;
 38     for(int i=head1[u]; i!=-1; i=E1[i].next)
 39         if(!mark1[E1[i].to])
 40             DFS1(E1[i].to);
 41     Et[cont1++]=u;
 42 }
 43 void DFS2(int u)
 44 {
 45     mark2[u]=1;
 46     num++;
 47     vis[u]=cont2;
 48     for(int i=head2[u]; i!=-1; i=E2[i].next)
 49     {
 50         if(!mark2[E2[i].to])
 51             DFS2(E2[i].to);
 52     }
 53 }
 54 int main()
 55 {
 56     int i,b;
 57     while(~scanf("%d",&n))
 58     {
 59         Init();
 60         for(i=1; i<=n; i++)
 61         {
 62             while(scanf("%d",&b),b)
 63             {
 64                 add(i,b);
 65                 e[k].to=i,e[k].next=b;
 66                 k++;
 67             }
 68         }
 69         for(i=1; i<=n; i++)
 70             if(!mark1[i])
 71                 DFS1(i);
 72         for(i=cont1-1; i>=1; i--)
 73         {
 74             if(!vis[Et[i]])
 75             {
 76                 num=0;
 77                 DFS2(Et[i]);
 78                 ss[cont2++]=num;
 79             }
 80         }
 81         for(i=1; i<=k; i++)
 82         {
 83             if(vis[e[i].to] != vis[e[i].next])
 84             {
 85                 out[vis[e[i].to]]++;
 86                 in[vis[e[i].next]]++;
 87             }
 88         }
 89         int sum;
 90         sum = num =0;
 91         for(i=1; i<cont2; i++)
 92         {
 93             if(!out[i])
 94                 sum++;
 95             if(!in[i])
 96                 num++;
 97         }
 98         if(cont2==2)
 99             printf("1\n0\n");
100         else
101         {
102             printf("%d\n",num);
103             printf("%d\n",max(sum,num));
104         }
105     }
106     return 0;
107 }

 

posted @ 2015-08-07 09:02  _Nestling  阅读(119)  评论(0)  编辑  收藏  举报