Hopcroft-Carp算法模板【二分图匹配】

来源：互联网发布：seo三人行编辑：程序博客网时间：2024/05/16 13:47
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)；
#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[N][N];int Mx[N];int My[N];int dx[N];int dy[N];bool used[N];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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