Strongly connected board, the first reduction point, and then only out of consideration for the zero point it may be the answer, but if the degree is more than one answer has 0 points or 0
I use disjoint-set whether to maintain the relationship on a chain
Code:
#include <bits/stdc++.h> #define int long long #define sc(a) scanf("%lld",&a) #define scc(a,b) scanf("%lld %lld",&a,&b) #define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define schar(a) scanf("%c",&a) #define pr(a) printf("%lld",a) #define fo(i,a,b) for(int i=a;i<b;++i) #define re(i,a,b) for(int i=a;i<=b;++i) #define rfo(i,a,b) for(int i=a;i>b;--i) #define rre(i,a,b) for(int i=a;i>=b;--i) #define prn() printf("\n") #define prs() printf(" ") #define mkp make_pair #define pii pair<int,int> #define pub(a) push_back(a) #define pob() pop_back() #define puf(a) push_front(a) #define pof() pop_front() #define fst first #define snd second #define frt front() #define bak back() #define mem0(a) memset(a,0,sizeof(a)) #define memmx(a) memset(a,0x3f3f,sizeof(a)) #define memmn(a) memset(a,-0x3f3f,sizeof(a)) #define debug #define db double #define yyes cout<<"YES"<<endl; #define nno cout<<"NO"<<endl; using namespace std; typedef vector<int> vei; typedef vector<pii> vep; typedef map<int,int> mpii; typedef map<char,int> mpci; typedef map<string,int> mpsi; typedef deque<int> deqi; typedef deque<char> deqc; typedef priority_queue<int> mxpq; typedef priority_queue<int,vector<int>,greater<int> > mnpq; typedef priority_queue<pii> mxpqii; typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii; const int maxn=500005; const int inf=0x3f3f3f3f3f3f3f3f; const int MOD=100000007; const db eps=1e-10; int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;} int lowbit(int x){return x&-x;} int max(int a,int b){return a>b?a:b;} int min(int a,int b){return a<b?a:b;} int mmax(int a,int b,int c){return max(a,max(b,c));} int mmin(int a,int b,int c){return min(a,min(b,c));} void mod(int &a){a+=MOD;a%=MOD;} bool chk(int now){} int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;} int ll(int p){return p<<1;} int rr(int p){return p<<1|1;} int mm(int l,int r){return (l+r)/2;} int lg(int x){if(x==0) return 1;return (int)log2(x)+1;} bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;} db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));} bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;} inline int read(){ char ch=getchar();int s=0,w=1; while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();} while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();} return s*w; } inline void write(int x){ if(x<0)putchar('-'),x=-x; if(x>9)write(x/10); putchar(x%10+48); } int n,m,x[maxn],y[maxn]; vei g[maxn],scc[maxn]; int c[maxn]; int dfn[maxn],low[maxn],timer=0,cnt=0; bool ins[maxn]; stack<int> s; int ind[maxn],outd[maxn],sz[maxn]; void tarjan(int x){ dfn[x]=low[x]=++timer; s.push(x);ins[x]=1; fo(i,0,g[x].size()){ int y=g[x][i]; if(!dfn[y]){ tarjan(y); low[x]=min(low[x],low[y]); } else if(ins[y]) low[x]=min(low[x],dfn[y]); } if(dfn[x]==low[x]){ cnt++; int y; do{ y=s.top(); s.pop(); ins[y]=0; c[y]=cnt,scc[cnt].pub(y); }while(x!=y); } } int f[maxn]; int _find(int x){ if(f[x]!=x) f[x]=_find(f[x]); return f[x]; } void _merge(int x,int y){ x=_find(x),y=_find(y); f[x]=y; } void build(){ re(i,1,cnt) f[i]=i; re(i,1,m){ if(c[x[i]]!=c[y[i]]) _merge(c[x[i]],c[y[i]]), outd[c[x[i]]]++,ind[c[y[i]]]++; } int ff=f[1]; re(i,1,cnt) if(_find(i)!=ff){cout<<0;return;} int ans=0; int tmp=0; re(i,1,cnt){ if(outd[i]==0) ans+=scc[i].size(),tmp++; } if(tmp>1) ans=0; cout<<ans; } signed main(){ ios_base::sync_with_stdio(0); cin.tie(0),cout.tie(0); cin>>n>>m; re(i,1,m){ cin>>x[i]>>y[i]; g[x[i]].pub(y[i]); } re(i,1,n) if(!dfn[i]) tarjan(i); build(); return 0; } /* 3 2 1 2 2 3 3 2 1 2 1 3 */