二分图最大匹配
其实就是求最大点权独立集,而最大点权独立集=总点数-最大匹配数。那么跑个传递闭包后建图求最大匹配即可。
代码:
#include<cctype>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 405
using namespace std;
struct edge{ int nxt,to; }ed[N*N<<1];
int n,m,t,ans,k,h[N],p[N],f[N];
bool a[N][N];
inline char readc(){
static char buf[100000],*l=buf,*r=buf;
if (l==r) r=(l=buf)+fread(buf,1,100000,stdin);
return l==r?EOF:*l++;
}
inline int _read(){
int x=0; char ch=readc();
while (!isdigit(ch)) ch=readc();
while (isdigit(ch)) x=(x<<3)+(x<<1)+(ch^48),ch=readc();
return x;
}
#define addedge(x,y) ed[++k]=(edge){h[x],y},h[x]=k
bool dfs(int x){
if (f[x]==t) return false; f[x]=t;
for (int i=h[x],v;i;i=ed[i].nxt)
if (!p[v=ed[i].to]||dfs(p[v]))
return p[v]=x,true;
return false;
}
int main(){
n=_read(),m=_read();
for (int i=1,x,y;i<=m;i++)
x=_read(),y=_read(),a[x][y]=true;
for (int m=1;m<=n;m++)
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
a[i][j]|=(a[i][m]&a[m][j]);
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
if (a[i][j]) addedge(i,n+j),addedge(n+j,i);
for (t=1;t<=n;t++) ans+=dfs(t);
return printf("%d",n-ans),0;
}