POJ2553 The Bottom of a Graph（强连通分量+缩点）

Posted 2020-06-08

题目是问，一个有向图有多少个点v满足∀w∈V:(v→w)⇒(w→v)。

把图的强连通分量缩点，那么答案显然就是所有出度为0的点。

用Tarjan找强连通分量：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 #define MAXN 5555
 6 #define MAXM 5555*5555
 7 struct Edge{
 8     int u,v,next;
 9 }edge[MAXM];
10 int NE,head[MAXN];
11 void addEdge(int u,int v){
12     edge[NE].u=u; edge[NE].v=v; edge[NE].next=head[u];
13     head[u]=NE++;
14 }
15 
16 int bn,belong[MAXN],stack[MAXN],top;
17 bool instack[MAXN];
18 int dn,dfn[MAXN],low[MAXN];
19 void dfs(int u){
20     dfn[u]=low[u]=++dn;
21     stack[++top]=u; instack[u]=1;
22     for(int i=head[u]; i!=-1; i=edge[i].next){
23         int v=edge[i].v;
24         if(dfn[v]==0){
25             dfs(v);
26             low[u]=min(low[u],low[v]);
27         }else if(instack[v]){
28             low[u]=min(low[u],dfn[v]);
29         }
30     }
31     if(dfn[u]==low[u]){
32         int v; ++bn;
33         do{
34             v=stack[top--];
35             instack[v]=0;
36             belong[v]=bn;
37         }while(u!=v);
38     }
39 }
40 
41 int deg[MAXN];
42 int main(){
43     int n,m,a,b;
44     while(~scanf("%d",&n) && n){
45         NE=0;
46         memset(head,-1,sizeof(head));
47         scanf("%d",&m);
48         while(m--){
49             scanf("%d%d",&a,&b);
50             addEdge(a,b);
51         }
52         top=dn=bn=0;
53         memset(dfn,0,sizeof(dfn));
54         memset(instack,0,sizeof(instack));
55         for(int i=1; i<=n; ++i){
56             if(dfn[i]==0) dfs(i);
57         }
58         memset(deg,0,sizeof(deg));
59         for(int i=0; i<NE; ++i){
60             int u=belong[edge[i].u],v=belong[edge[i].v];
61             if(u==v) continue;
62             ++deg[u];
63         }
64         bool first=1;
65         for(int i=1; i<=n; ++i){
66             if(deg[belong[i]]==0){
67                 if(first) first=0;
68                 else putchar(‘ ‘);
69                 printf("%d",i);
70             }
71         }
72         putchar(‘\n‘);
73     }
74     return 0;
75 }