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Hopcroft-Carp算法模板【二分图匹配】

2015-06-15 20:17 411 查看
模板://hdu 2063

Hopcroft-Carp 时间复杂度为 O(sqrt(V)*E);

而匈牙利算法为 O(V*E);

[code]#include <stdio.h>
#include <ctime>
#include <math.h>
#include <limits.h>
#include <complex>
#include <string>
#include <functional>
#include <iterator>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <bitset>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <iostream>
#include <ctime>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <time.h>
#include <ctype.h>
#include <string.h>
#include <assert.h>

using namespace std;

const int N = 1005;
const int INF = 1 << 28;

int g

;
int Mx
;
int My
;
int dx
;
int dy
;
bool used
;

int Nx, Ny, dis;

bool searchP()
{
    dis = INF;
    int i, v, u;
    std::queue<int> Q;

    memset(dx, -1, sizeof(dx));
    memset(dy, -1, sizeof(dy));
    for (i = 0; i<Nx; i++)
    {
        if (Mx[i] == -1)
        {
            Q.push(i);
            dx[i] = 0;
        }
    }
    while (!Q.empty())
    {
        u = Q.front();
        Q.pop();
        if (dx[u]>dis) break;
        for (v = 0; v<Ny; v++)
        {
            if (g[u][v] && dy[v] == -1)
            {
                dy[v] = dx[u] + 1;
                if (My[v] == -1) dis = dy[v];
                else
                {
                    dx[My[v]] = dy[v] + 1;
                    Q.push(My[v]);
                }
            }
        }
    }
    return dis != INF;
}

bool DFS(int u)
{
    int v;
    for (v = 0; v<Ny; v++)
    {
        if (g[u][v] && !used[v] && dy[v] == dx[u] + 1)
        {
            used[v] = true;
            if (My[v] != -1 && dy[v] == dis) continue;
            if (My[v] == -1 || DFS(My[v]))
            {
                My[v] = u;
                Mx[u] = v;
                return true;
            }
        }
    }
    return false;
}

int Hungary()
{
    int u;
    int ret = 0;
    memset(Mx, -1, sizeof(Mx));
    memset(My, -1, sizeof(My));
    while (searchP())
    {
        memset(used, false, sizeof(used));
        for (u = 0; u<Nx; u++)
            if (Mx[u] == -1 && DFS(u))  ret++;
    }
    return ret;
}

int main()
{
    int k, u, v;
    while (~scanf("%d", &k) ,k)
    {
        scanf("%d%d", &Nx, &Ny);
        memset(g, 0, sizeof(g));
        Ny = Nx>Ny ? Nx : Ny;
        while (k--)
        {
            scanf("%d%d", &u, &v);
            u--; v--;
            g[u][v] = 1;
        }
        int ans = Hungary();
        printf("%d\n", ans);
    }
    return 0;
}
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