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-9;
inline int sgn(DB x){if(abs(x)<EPS)return 0;return(x>0)?1:-1;}
const DB Pi=acos(-1.0);

inline int gint()
  {
        int res=0;bool neg=0;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z=='-'){neg=1;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
        return (neg)?-res:res; 
    }
inline LL gll()
  {
      LL res=0;bool neg=0;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z=='-'){neg=1;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
        return (neg)?-res:res; 
    }

const int maxN=100;

int N,M;
int mp[maxN+10][maxN+10];

int first[maxN+100],now;
struct Tedge{int v,next;}edge[maxN*maxN+10000];
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+100];
int form[maxN+100];

inline int DFS(int u)
  {
      int i,v;
      vis[u]=1;
      for(i=first[u],v=edge[i].v;i!=-1;i=edge[i].next,v=edge[i].v)
        if(!form[v] || (!vis[form[v]] && DFS(form[v])))
          {
              form[v]=u;
              return 1;
          }
      return 0;
  }

int main()
  {
      freopen("bzoj1143.in","r",stdin);
      freopen("bzoj1143.out","w",stdout);
      int i,j,k;
      N=gint();M=gint();
      re(i,1,M){int u=gint(),v=gint();mp[u][v]=1;}
      re(k,1,N)re(i,1,N)re(j,1,N)if(i!=k && j!=k && i!=j && mp[i][k] && mp[k][j]) mp[i][j]=1;
      mmst(first,-1);now=-1;
      re(i,1,N)re(j,1,N)if(mp[i][j])addedge(i,j);
      ans=0;
      re(i,1,N)
        {
            re(j,1,N)vis[j]=0;
            ans+=DFS(i);
        }
        printf("%d\n",N-ans);
        return 0;
    }
View Code

 

posted @ 2015-07-15 21:02  maijing  阅读(175)  评论(0编辑  收藏  举报