加上多少条边能成为强连通
#include<bits/stdc++.h> using namespace std; //input by bxd #define rep(i,a,b) for(int i=(a);i<=(b);i++) #define repp(i,a,b) for(int i=(a);i>=(b);--i) #define RI(n) scanf("%d",&(n)) #define RII(n,m) scanf("%d%d",&n,&m) #define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k) #define RS(s) scanf("%s",s); #define ll long long #define pb push_back #define REP(i,N) for(int i=0;i<(N);i++) #define CLR(A,v) memset(A,v,sizeof A) ////////////////////////////////// #define inf 0x3f3f3f3f const int N=20000+5; int head[10*N],pos; struct Edge { int to,nex; }edge[10*N]; void add(int a,int b) { edge[++pos].nex=head[a]; head[a]=pos; edge[pos].to=b; } int dfn[N],low[N],tot,vis[N],ind,Stack[N],cnt,belong[N]; int in[N],out[N]; int init() { CLR(head,0); CLR(dfn,0); CLR(low,0); CLR(vis,0); CLR(Stack,0); CLR(belong,0); CLR(in,0); CLR(out,0); cnt=ind=tot=pos=0; } void tarjan(int x) { dfn[x]=low[x]=++tot; Stack[++ind]=x; vis[x]=1; for(int i=head[x];i;i=edge[i].nex) { int v=edge[i].to; if(!dfn[v]) { tarjan(v);low[x]=min(low[x],low[v]); } else if(vis[v]) low[x]=min(low[x],low[v]); } if(low[x]==dfn[x]) { cnt++; int v; do { v=Stack[ind--]; vis[v]=0; belong[v]=cnt; } while(x!=v); } } int main() { int n,m; while(RII(n,m)==2) { init(); rep(i,1,m) { int a,b;RII(a,b); add(a,b); } rep(i,1,n) if(!dfn[i]) tarjan(i); int Cnt=0; rep(i,1,n) { int u=belong[i]; for(int j=head[i];j;j=edge[j].nex) { int v=belong[edge[j].to]; if(u!=v) out[u]++,in[v]++; } } int q=0,w=0; rep(i,1,cnt) { if(!in[i])q++; if(!out[i])w++; } if(cnt==1) cout<<0<<endl; else cout<<max(q,w)<<endl; } return 0; }
原文:https://www.cnblogs.com/bxd123/p/10798197.html