Cod sursa(job #1389672)

Utilizator AdrianaMAdriana Moisil AdrianaM Data 16 martie 2015 15:19:44
Problema Cuplaj maxim in graf bipartit Scor 50
Compilator cpp Status done
Runda Arhiva educationala Marime 2.21 kb
#include <fstream>
#include <vector>
#include <queue>
#define INF 0x3f3f3f3f
#define nodv g[nod][i]
using namespace std;

ifstream is("cuplaj.in");
ofstream os("cuplaj.out");

int n, m, k;
vector<short> t;
vector<vector<short> > g, cap;

void Read();
bool Bfs();

int main()
{
    Read();
    t = vector<short>(n + m + 2);
    short answ = 0, fmin, nod = n + m + 1;
    while ( Bfs() )
        for ( size_t i = 0; i < g[nod].size(); ++i )
        {
            if ( !t[nodv] || cap[nodv][nod] <= 0 )
                continue;
            t[nod] = nodv;
            fmin = INF;
            for ( int x = nod; x; x = t[x] )
                fmin = min(fmin, cap[t[x]][x]);
            if ( !fmin )
                continue;
            for ( int x = nod; x; x = t[x] )
            {
                cap[t[x]][x] -= fmin;
                cap[x][t[x]] += fmin;
            }
            answ += fmin;
        }
    os << answ << "\n";
    for ( int i = 1; i <= n && answ; ++i )
        for ( int j = n + 1; j <= n + m && answ; ++j )
            if ( cap[j][i] )
            {
                os << i << " " << j - n << "\n";
                --answ;
            }
    is.close();
    os.close();
    return 0;
}

bool Bfs()
{
    t.assign(n + m + 2, 0);
    queue<short> q;
    q.push(0);
    int nod;
    while ( !q.empty() )
    {
        nod = q.front();
        q.pop();
        if ( nod == n + m + 1 )
            continue;
        for ( size_t i = 0; i < g[nod].size(); ++i )
            if ( nodv && !t[nodv] && cap[nod][nodv] > 0 )
            {
                t[nodv] = nod;
                q.push(nodv);
            }
    }
    return t[n + m + 1];
}

void Read()
{
    is >> n >> m >> k;
    g = vector<vector<short> >(n + m + 2);
    cap = vector<vector<short> >(n + m + 2, vector<short>(n + m + 2));
    int n1, n2;
    while ( k-- )
    {
        is >> n1 >> n2;
        g[n1].push_back(n + n2);
        g[n + n2].push_back(n1);
        cap[n1][n + n2] = 1;
    }
    for ( int i = 1; i <= n; ++i )
    {
        g[0].push_back(i);
        g[i].push_back(0);
        cap[0][i] = 1;
    }
    for (int i = n + 1; i <= n + m; ++i )
    {
        g[i].push_back(n + m + 1);
        g[n + m + 1].push_back(i);
        cap[i][n + m + 1] = 1;
    }
}