Cod sursa(job #1376879)

Utilizator corul_barbatescUNIBUC Kira96 lockmihai corul_barbatesc Data 5 martie 2015 19:18:19
Problema Cuplaj maxim in graf bipartit Scor 100
Compilator cpp Status done
Runda Arhiva educationala Marime 1.94 kb
/*
    Those without power, seek us!
    Those with power, fear us!
    We are the Order of the Black Knights!
*/

#include<fstream>
#include<cstdio>
#include<map>
#include<set>
#define FIT(a,b) for(vector<int >::iterator a=b.begin();a!=b.end();a++)
#define FITP(a,b) for(vector<pair<int,int> >::iterator a=b.begin();a!=b.end();a++)
#define RIT(a,b) for(vector<int>::reverse_iterator a=b.end();a!=b.begin();++a)
#include<stack>
#define ROF(a,b,c) for(int a=b;a>=c;--a)
#include<vector>
#include<algorithm>
#define FOR(a,b,c) for(int a=b;a<=c;++a)
#define REP(a,b) for(register int a=0;a<b;++a)
#include<cstring>
#include<ctime>
#include<bitset>
#include<cmath>
#include<iomanip>
#include<set>
#define f cin
#define g cout
#include<queue>
#define debug cerr<<"OK";
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ll long long
#define ull unsigned int
#define mod 1000000009LL
#define SQR 350
#define inf 1<<30
#define div fdasfasd
#define hash dsafdsfds
#define od 100003
#define mod  1999999973LL
#define DIM 60010000
#define base 256
#define bas 255
#define N 100100
using namespace std;
ifstream f("cuplaj.in");
ofstream g("cuplaj.out");
/*
    int dx[]={0,0,0,1,-1};
    int dy[]={0,1,-1,0,0};
*/
vector<int> v[N];
int L[N],R[N],n,x,y,ok,sol,viz[N],m,p;
int cuplaj(int x)
{
    if(viz[x])
        return 0;
    viz[x]=1;
    FIT(it,v[x])
    if(!R[*it])
    {
        L[x]=*it;
        R[*it]=x;
        return 1;
    }

    FIT(it,v[x])
    if(cuplaj(R[*it]))
    {
        L[x]=*it;
        R[*it]=x;
        return 1;
    }
    return 0;
}
int main ()
{
    f>>n>>m>>p;
    FOR(i,1,p)
    {
        f>>x>>y;
        v[x].pb(y);
    }
    ok=1;
    while(ok)
    {
        ok=0;
        memset(viz,0,sizeof(viz));
        FOR(i,1,n)
        if(!L[i]&&cuplaj(i))
        {
            ++sol;
            ok=1;
        }
    }
    g<<sol<<"\n";
    FOR(i,1,n)
    if(L[i])
        g<<i<<" "<<L[i]<<"\n";
    return 0;
}