二分图最大匹配 -- 匈牙利算法


Algorithm.( Augmenting Path Algorithm )


Input:
    An X-Y bigraph G, a matching M in G,
    and the set U of M-unsaturated vertices in X.
        
Idea:
    Explore M-alternating paths form U,
    letting S ⊆ X and T ⊆ Y be the sets of vertices reached.
    Marks vertices of S that have been explored for path extensions.
    As a vertex is reached, record the vertex from which it is reached.
        
Initialization:
    S = U and T = ∅


Iteration:
    If S has no unmarked vertex,
    stop and report T ∪ ( X - S ) as a minimun cover and M as a maximum matching.
    Otherwise, select an unmarked x ∈ S, consider each y ∈ N( x ) such that xy ∉ M,
    if y is unsaturated, terminate and report an M-augmenting path from U to y.
    Otherwise, y is matched to some w ∈ X by M.
    In this case, include y in T ( reached from x ) and include w in S ( reached from y ).
    After exploring all such edges incident to x, mark x and iterate.


#include <iostream>
#include <cstring>
using namespace std;

#define NODE_SIZE 100

bool bigraph[NODE_SIZE][NODE_SIZE];
int parent[NODE_SIZE];
bool visit[NODE_SIZE];

int num = 0;
int nodes1, nodes2, edges;


bool find_augmenting_path( int start_node ){

    for( int end_node = 1; end_node <= nodes2; ++end_node ){

        if( bigraph[start_node][end_node] && visit[end_node] == false ){
            visit[end_node] = true;

            if( parent[end_node] == 0 ||
               find_augmenting_path( parent[end_node] ) ){
                parent[end_node] = start_node;
                return true;
            }
        }
    }
    return false;
}

int main(){

    memset( bigraph, false, sizeof( bigraph ) );
    memset( visit, false, sizeof( visit ) );
    memset( parent, 0, sizeof( parent ) );

    cin >> nodes1 >> nodes2 >> edges;

    for( int i = 1; i <= edges; ++i ){
        int start_node, end_node;
        cin >> start_node >> end_node;
        bigraph[start_node][end_node] = true;
    }

    for( int node = 1; node <= nodes1; ++node ){
        memset( visit, false, sizeof( visit ) );
        if( find_augmenting_path( node ) )
            num++;
    }

    cout << num << endl;
    return 0;
}


简单应用:

在 raw * col 的棋盘上,有些格子不能放。用1 * 2 的方块铺棋盘。问能不能铺满?

比方:

思路:

对每种格子着色,黑白相间,不能放的格子除。

然后白色的格子为一个部集,黑色的格子也为一个部集。两个部集进行最大匹配,

若最后匹配数目等于黑色或者白色格子的总数,即为可行;否则,不可行。


法1:

// 1143MS
#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

#define CHESSBOARD_SIZE 50
#define GRAPH_SIZE 1100

bool chessboard[CHESSBOARD_SIZE][CHESSBOARD_SIZE];
bool bigraph[GRAPH_SIZE][GRAPH_SIZE];
bool visit[GRAPH_SIZE];
int parent[GRAPH_SIZE];
int raws, cols, holes;

bool find_augmenting_path( int source ){
    for( int target = 1; target <= raws * cols; ++target ){
        if( bigraph[source][target] && !visit[target] ){
            visit[target] = true;
            if( parent[target] == 0 || find_augmenting_path( parent[target] ) ){
                parent[target] = source;
                return true;
            }
        }
    }
    return false;
}

int maximum_matching(){
    int ans = 0;
    for( int i = 1; i <= raws * cols; ++i ){
        memset( visit, false, sizeof( visit ) );
        if( find_augmenting_path( i ) )
            ans++;
    }
    return ans;
}

int main(){

    memset( chessboard, false, sizeof( chessboard ) );
    memset( bigraph, false, sizeof( bigraph ) );
    memset( visit, true, sizeof( visit ) );
    memset( parent, 0, sizeof( parent ) );

    cin >> raws >> cols >> holes;
    int num = raws * cols;

    for( int i = 1; i <= holes; ++i ){
        int raw, col;
        cin >> col >> raw;
        chessboard[raw][col] = true;
        num--;
    }

    for( int raw = 1; raw <= raws; ++raw ){
        for( int col = 1; col <= cols; ++col ){
            if( chessboard[raw][col] )
                continue;
            int p1 = cols * ( raw - 1 ) + col;
            if( raw > 1 && chessboard[raw - 1][col] == false ){
                int p2 = p1 - cols;
                bigraph[p1][p2] = true;
            }
            if( raw < raws && chessboard[raw + 1][col]== false ){
                int p2 = p1 + cols;
                bigraph[p1][p2] = true;
            }
            if( col > 1 && chessboard[raw][col - 1] == false ){
                int p2 = p1 - 1;
                bigraph[p1][p2] = true;
            }
            if( col < cols && chessboard[raw][col + 1] == false ){
                int p2 = p1 + 1;
                bigraph[p1][p2] = true;
            }
        }
    }
    int ans = maximum_matching();
    if( ans == num )
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
    return 0;
}

法2:

// 725k 32MS
#include <iostream>
#include <vector>
#include <cstring>
using namespace std;

#define NODE_SIZE 50

struct Position{
    Position(){
        raw = -1;
        col = -1;
    };
    Position( int r, int c ){
        raw = r;
        col = c;
    };
    int raw;
    int col;
};

enum State { INIT, HOLE, BLACK, WHITE };
State chessboard[NODE_SIZE][NODE_SIZE];

Position parent[NODE_SIZE * NODE_SIZE];
bool visit[NODE_SIZE * NODE_SIZE];
int raws, cols, holes;

int direction[5][3] = {
    { 0, 0, 0 },
    { 0, 1, 0 },
    { 0, 0, 1 },
    { 0, -1, 0 },
    { 0, 0, -1 }
};

bool find_augmenting_path( Position source ){
    for( int i = 1; i <= 4; ++i ){
        int dx = direction[i][1];
        int dy = direction[i][2];
        int next_raw = source.raw + dx;
        int next_col = source.col + dy;

        if( next_raw >= 1 &&
            next_raw <= raws &&
            next_col >= 1 &&
            next_col <= cols ){

            int one_dim_pos = cols * ( next_raw - 1 ) + next_col;
            if( chessboard[next_raw][next_col] == WHITE && !visit[one_dim_pos] ){

                visit[one_dim_pos] = true;
                if( ( parent[one_dim_pos].raw == -1 &&
                    parent[one_dim_pos].col == -1 ) ||
                    find_augmenting_path( parent[one_dim_pos] ) ){

                    parent[one_dim_pos].raw = source.raw;
                    parent[one_dim_pos].col = source.col;
                    return true;
                }
            }
        }
    }
    return false;
}

int main(){

    cin >> raws >> cols >> holes;
    vector< Position > black_rects;
    vector< Position > white_rects;

    for( int raw = 1; raw <= raws; ++raw ){
        for( int col = 1; col <= cols; ++col ){
            chessboard[raw][col] = INIT;
        }
    }
    memset( visit, false, sizeof( visit ) );

    int rects = raws * cols - holes;

    for( int i = 1; i <= holes; ++i ){
        int col, raw;
        cin >> col >> raw;
        chessboard[raw][col] = HOLE;
    }

    for( int raw = 1; raw <= raws; ++raw ){
        for( int col = 1; col <= cols; ++col ){
            if( chessboard[raw][col] == HOLE )
                continue;
            if( ( col % 2 && raw % 2 ) ||
                ( ( raw % 2 == 0 ) && ( col % 2 == 0 ) ) ){
                chessboard[raw][col] = BLACK;
                black_rects.push_back( Position( raw, col ) );
            }
            else{
                chessboard[raw][col] = WHITE;
                white_rects.push_back( Position( raw, col ) );
            }
        }
    }

    const int black_rects_num = black_rects.size();
    const int white_rects_num = white_rects.size();

    if( black_rects_num == 0 ||
            rects % 2 != 0 ||
            black_rects_num != white_rects_num ){
        cout << "NO" << endl;
    }
    else{
        int ans = 0;
        for( int i = 0; i < black_rects_num; ++i ){
            memset( visit, false, sizeof( visit ) );
            if( find_augmenting_path( black_rects[i] ) )
                ans++;
        }
        if( ans == black_rects_num ){
            cout << "YES" << endl;
        }
        else{
            cout << "NO" << endl;
        }
    }
    return 0;
}


POJ 3692

建反图,求最大独立集

#include <iostream>
#include <cstring>
using namespace std;

#define MAX_SIZE 210

bool bigraph[MAX_SIZE][MAX_SIZE];
bool visits[MAX_SIZE];
int parents[MAX_SIZE];
int B, G, M;

bool find_augmenting_path( int source ){
    for( int target = 1; target <= B; ++target ){
        if( bigraph[source][target] && !visits[target] ){
            visits[target] = true;
            if( parents[target] == 0 || find_augmenting_path( parents[target] ) ){
                parents[target] = source;
                return true;
            }
        }
    }
    return false;
}

int maximum_matching(){
    int ans = 0;
    for( int source = 1; source <= G; ++source ){
        memset( visits, false, sizeof( visits ) );
        if( find_augmenting_path( source ) )
            ans++;
    }
    return ans;
}

int main(){

    int nCase = 1;

    while( true ){
        memset( bigraph, true, sizeof( bigraph ) );
        memset( parents, 0, sizeof( parents ) );
        memset( visits, false, sizeof( visits ) );
        cin >> G >> B >> M;

        if( G == 0 && B == 0 && M == 0 )
            break;

        for( int i = 1; i <= M; ++i ){
            int u, v;
            cin >> u >> v;
            bigraph[u][v] = false;
        }
        int max_matching = maximum_matching();
        int ans = G + B - max_matching;
        cout << "Case " << nCase << ": " << ans << endl;
        nCase++;
    }
    return 0;
}


posted @ 2017-04-15 18:03  zhchoutai  阅读(223)  评论(0编辑  收藏  举报