bzoj 1191

http://www.lydsy.com/JudgeOnline/problem.php?id=1191

二分+二分图匹配。

首先二分可以答对前mid道题,然后做二分图。

左边是题目,右边是锦囊。

做匈牙利即可。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<utility>
#include<set>
#include<bitset>
#include<vector>
#include<ctime>

using namespace std;

typedef long long LL;
typedef double DB;
typedef pair<int,int> PII;

#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

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 dblcmp(DB x){if(abs(x)<EPS)return 0;return(x>0)?1:-1;}

inline void SetOpen(string s)
  {
        freopen((s+".in").c_str(),"r",stdin);
        freopen((s+".out").c_str(),"w",stdout);
    }

inline int Getin_Int()
  {
        int res=0,flag=1;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z=='-'){flag=-flag;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
        return res*flag;
    }
inline LL Getin_LL()
  {
        LL res=0,flag=1;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z=='-'){flag=-flag;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
        return res*flag;
    }

const int maxN=1000;
const int maxM=1000;

int N,M;
PII a[maxM+10];
int l,r,mid;

int first[maxN+10],now=-1;
struct Tedge{int v,next;}edge[2*maxM+100];

inline void addedge(int u,int v)
  {
        now++;
        edge[now].v=v;
        edge[now].next=first[u];
        first[u]=now;
    }

int form[maxM+10];
int vis[maxM+10];
inline int match(int x)
  {
        int i,v;
        for(i=first[x],v=edge[i].v;i!=-1;i=edge[i].next,v=edge[i].v)
          if(!vis[v])
            {
                    vis[v]=1;
                    if(form[v]==-1 || match(form[v]))
                      {
                            form[v]=x;
                            return 1;
                        }
                }
        return 0;
    }

inline int check(int mid)
  {
        int i;
        mmst(first,-1);now=-1;
        re(i,1,mid){addedge(a[i].fi,i);addedge(a[i].se,i);}
        mmst(form,-1);
        int tot=0;
        re(i,1,N)
          {
                mmst(vis,0);
                if(match(i))tot++;
            }
        return tot==mid;
    }

int main()
  {
        int i;
        SetOpen("hero");
        N=Getin_Int();M=Getin_Int();
        re(i,1,M)a[i].fi=Getin_Int()+1,a[i].se=Getin_Int()+1;
        l=1,r=M;
        while(l<=r)
          {
                int mid=(l+r)/2;
                if(check(mid))l=mid+1; else r=mid-1;
            }
        printf("%d
",r);
        /*int tmp=check(r);
        re(i,1,r)printf("%d
",form[i]-1);*/
        return 0;
    }
View Code
原文地址:https://www.cnblogs.com/maijing/p/4747579.html