题解:
最小链覆盖=最长反链
ans=n-Maxflow()
#include<iostream> #include<cstdio> #include<cstring> #include<vector> #include<queue> using namespace std; const int maxn=300; const int oo=100000000; int n,m; int g[maxn][maxn]; struct Edge{ int from,to,cap,flow; }; vector<int>G[maxn]; vector<Edge>edges; void Addedge(int x,int y,int z){ Edge e; e.from=x;e.to=y;e.cap=z;e.flow=0; edges.push_back(e); e.from=y;e.to=x;e.cap=0;e.flow=0; edges.push_back(e); int c=edges.size(); G[x].push_back(c-2); G[y].push_back(c-1); } int s,t; int vis[maxn]; int d[maxn]; queue<int>q; int Bfs(){ memset(vis,0,sizeof(vis)); vis[s]=1;d[s]=0;q.push(s); while(!q.empty()){ int x=q.front();q.pop(); for(int i=0;i<G[x].size();++i){ Edge e=edges[G[x][i]]; if((!vis[e.to])&&(e.cap>e.flow)){ vis[e.to]=1; d[e.to]=d[x]+1; q.push(e.to); } } } return vis[t]; } int cur[maxn]; int Dfs(int x,int a){ if((x==t)||(a==0))return a; int nowflow=0,f=0; for(int i=cur[x];i<G[x].size();++i){ cur[x]=i; Edge e=edges[G[x][i]]; if((d[x]+1==d[e.to])&&((f=Dfs(e.to,min(a,e.cap-e.flow)))>0)){ nowflow+=f; a-=f; edges[G[x][i]].flow+=f; edges[G[x][i]^1].flow-=f; if(a==0)break; } } return nowflow; } int Maxflow(){ int flow=0; while(Bfs()){ memset(cur,0,sizeof(cur)); flow+=Dfs(s,oo); } return flow; } int main(){ scanf("%d%d",&n,&m); for(int i=1;i<=m;++i){ int x,y; scanf("%d%d",&x,&y); g[x][y]=1; } for(int k=1;k<=n;++k){ for(int i=1;i<=n;++i){ for(int j=1;j<=n;++j){ g[i][j]=(g[i][j]|(g[i][k]&g[k][j])); } } } // for(int i=1;i<=n;++i){ // for(int j=1;j<=n;++j){ // cout<<g[i][j]<<' '; // } // cout<<endl; // } for(int i=1;i<=n;++i){ for(int j=1;j<=n;++j){ if(g[i][j]){ Addedge(i,n+j,1); } } } s=n+n+1;t=s+1; for(int i=1;i<=n;++i){ Addedge(s,i,1); Addedge(i+n,t,1); } cout<<n-Maxflow()<<endl; return 0; }