强连通分量：
首先tarjan缩点重构图
之后，若出度为0的点仅有一个，那么答案即为该点代表的强连通分量中点的个数
否则，答案为0
1 #include<cstdio> 2 #include<cstring> 3 using namespace std; 4 const int N=10010,M=50010; 5 const int novis=-1,over=1,nowvis=0; 6 int size,head[M],next[M],to[M]; 7 int head2[M],next2[M],to2[M]; 8 int low[N],dfn[N],flag[N],color[N],que[N],sum[N], 9 top,cnt,n,m,sig; 10 void tarjan(),uni(int,int),dfs(int),rebuild(); 11 int find(),min(int,int); 12 int main(){ 13 int ans,x,y; 14 size=0; 15 memset(head,0,sizeof(head)); 16 memset(head2,0,sizeof(head2)); 17 scanf("%d %d",&n,&m); 18 for (int i=1;i<=m;i++){ 19 scanf("%d %d",&x,&y); 20 uni(x,y); 21 } 22 tarjan(); 23 rebuild(); 24 ans=find(); 25 printf("%d",ans); 26 return 0; 27 } 28 void uni(int x,int y){ 29 size++; 30 next[size]=head[x]; 31 head[x]=size; 32 to[size]=y; 33 } 34 void tarjan(){ 35 memset(flag,novis,sizeof(flag)); 36 memset(color,0,sizeof(color)); 37 memset(sum,0,sizeof(sum)); 38 sig=cnt=top=0; 39 for (int i=1;i<=n;i++) 40 if (flag[i]==novis) dfs(i); 41 } 42 void dfs(int x){ 43 que[++top]=x; 44 flag[x]=nowvis; 45 low[x]=dfn[x]=++sig; 46 for (int e=head[x];e;e=next[e]){ 47 int v=to[e]; 48 if (flag[v]==novis){ 49 dfs(v); 50 low[x]=min(low[x],low[v]); 51 } 52 else if (flag[v]==nowvis) 53 low[x]=min(low[x],dfn[v]); 54 } 55 if (low[x]==dfn[x]){ 56 int t;cnt++; 57 do{ 58 t=que[top--]; 59 color[t]=cnt; 60 flag[t]=over; 61 sum[cnt]++; 62 }while (t!=x); 63 } 64 } 65 void rebuild(){ 66 size=0; 67 for (int u=1;u<=n;u++){ 68 for (int e=head[u];e;e=next[e]){ 69 int v=to[e]; 70 if (color[u]!=color[v]){ 71 size++; 72 next2[size]=head2[color[u]]; 73 head2[color[u]]=size; 74 to2[size]=color[v]; 75 } 76 } 77 } 78 } 79 int find(){ 80 int ans=0; 81 for (int i=1;i<=cnt;i++) 82 if (!head2[i]){ 83 if (ans) return 0; 84 else ans=sum[i]; 85 } 86 return ans; 87 } 88 int min(int x,int y){ 89 return x<y?x:y; 90 }
