bzoj1143

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=1143

首先用传递闭包，知道一个点是否可以到达另一个点，即mp[i][j]==1表示i可以到j；mp[i][j]==0表示i不可以到j。

然后变成求有向无环图的最大独立集。

这是个经典问题，要变成二分图。

将每个点拆成两个点x和y

如果有边i->j，那么连边ix->jy。

然后求二分图的最大匹配，N-最大匹配就是答案。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<utility>
#include<set>
#include<bitset>
#include<vector>
#include<functional>
#include<deque>
#include<cctype>
#include<climits>
#include<complex>
//#include<bits/stdc++.h>适用于CF,UOJ，但不适用于poj
 
using namespace std;
 
typedef long long LL;
typedef double DB;
typedef pair<int,int> PII;
typedef complex<DB> CP;
 
#define mmst(a,v) memset(a,v,sizeof(a))
#define mmcy(a,b) memcpy(a,b,sizeof(a))
#define re(i,a,b)  for(i=a;i<=b;i++)
#define red(i,a,b) for(i=a;i>=b;i--)
#define fi first
#define se second
#define m_p(a,b) make_pair(a,b)
#define SF scanf
#define PF printf
#define two(k) (1<<(k))
 
template<class T>inline T sqr(T x){return x*x;}
template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;}
template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}
 
const DB EPS=1e-;
inline int sgn(DB x){if(abs(x)<EPS)return ;return(x>)?:-;}
const DB Pi=acos(-1.0);
 
inline int gint()
  {
        int res=;bool neg=;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return ;
        if(z=='-'){neg=;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*+z-'',z=getchar());
        return (neg)?-res:res;
    }
inline LL gll()
  {
      LL res=;bool neg=;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return ;
        if(z=='-'){neg=;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*+z-'',z=getchar());
        return (neg)?-res:res;
    }
 
const int maxN=;
 
int N,M;
int mp[maxN+][maxN+];
 
int first[maxN+],now;
struct Tedge{int v,next;}edge[maxN*maxN+];
int ans;
 
inline void addedge(int u,int v)
  {
      now++;
      edge[now].v=v;
      edge[now].next=first[u];
      first[u]=now;
  }
 
int vis[maxN+];
int form[maxN+];
 
inline int DFS(int u)
  {
      int i,v;
      vis[u]=;
      for(i=first[u],v=edge[i].v;i!=-;i=edge[i].next,v=edge[i].v)
        if(!form[v] || (!vis[form[v]] && DFS(form[v])))
          {
              form[v]=u;
              return ;
          }
      return ;
  }
 
int main()
  {
      freopen("bzoj1143.in","r",stdin);
      freopen("bzoj1143.out","w",stdout);
      int i,j,k;
      N=gint();M=gint();
      re(i,,M){int u=gint(),v=gint();mp[u][v]=;}
      re(k,,N)re(i,,N)re(j,,N)if(i!=k && j!=k && i!=j && mp[i][k] && mp[k][j]) mp[i][j]=;
      mmst(first,-);now=-;
      re(i,,N)re(j,,N)if(mp[i][j])addedge(i,j);
      ans=;
      re(i,,N)
        {
            re(j,,N)vis[j]=;
            ans+=DFS(i);
        }
        printf("%d\n",N-ans);
        return ;
    }