LOJ6226 「網絡流 24 題 - 22」 騎士共存 二分圖墜大點獨立集

大家都很強， 可與之共勉 。

題意：
  這道題有圖呃

題解：
  二分圖，然後……同方格取數，只不過把每個點點權看爲1 ，只能是左邊的點向右邊連邊。

  數組開小了RERERERERERERERERERRERERERERERER

# include <bits/stdc++.h>

# define oo 0x3f3f3f3f

# define N 40040

class Network  {
private :
    struct edge  {
        int to, w, nxt ;
        edge ( ) {        }
        edge ( int to, int w, int nxt ) : to ( to ), w ( w ), nxt ( nxt ) {        }  
    } g [250010 << 1] ;

    int head [N], cur [N], ecnt ;
    int S, T , dep [N] ;

    inline int dfs ( int u, int a )  {
        if ( u == T || ! a )  return a ;
        int flow = 0, v, f ;
        for ( int& i = cur [u] ; i ; i = g [i].nxt )  {
            v = g [i].to ;
            if ( dep [v] == dep [u] + 1 )  {
                f = dfs ( v, std :: min ( g [i].w, a - flow ) ) ;
                g [i].w -= f, g [i ^ 1].w += f ;
                flow += f ;
                if ( a == flow )  return a ;
            }
        }
        if ( ! flow )  dep [u] = -1 ;
        return flow ;
    }

    inline bool bfs ( int S, int T )  {
        static std :: queue < int > q ;
        memset ( dep, 0, sizeof ( int ) * ( T + 1 ) ) ;
        dep [S] = 1 ;
        q.push ( S ) ;
        while ( ! q.empty ( ) )  {
            int u = q.front ( ) ;  q.pop ( ) ;
            for ( int i = head [u] ; i ; i = g [i].nxt )  {
                int& v = g [i].to ;
                if ( g [i].w &&  ! dep [v] )  {
                    dep [v] = dep [u] + 1 ;
                    q.push ( v ) ;
                }
            }
        }
        return dep [T] ;
    }
public :
    Network ( )  {    ecnt = 1 ; }

    inline void add_edge ( int u, int v, int w )  {
        g [++ ecnt] = edge ( v, w, head [u] ) ;     head [u] = ecnt ;
        g [++ ecnt] = edge ( u, 0, head [v] ) ;     head [v] = ecnt ;
    }

    inline void clear ( )  {
        ecnt = 1 ;
        memset ( head, 0, sizeof head ) ;

    }
    inline int dinic ( int S, int T )  {
        this -> S = S, this -> T = T ;
        static int rt = 0 ;
        while ( bfs ( S, T ) )    {
            memcpy ( cur, head, sizeof ( int ) * ( T + 1 ) ) ; 
            rt += dfs ( S, 0x3f3f3f3f ) ;
        }
        return rt ;
    }
} Lazer ;

# define id(i,j) ((i-1)*n+j)

bool map [250] [250] ;

const int dx [] = { 1, 1, -1, -1, 2, 2, -2, -2 } ;
const int dy [] = { 2, -2, 2, -2, 1, -1, 1, -1 } ;

int main ( )  {
    int n, m ;
    scanf ( "%d%d", & n, & m ) ;
    int sum = n * n ;
    const int S = n * n + 1, T = n * n + 2 ;
    for ( int i = 1 ; i <= m ; ++ i )  {
        int x, y ;
        scanf ( "%d%d", & x, & y ) ;
        map [x] [y] = 1 ;
        -- sum ;
    }
    for  ( int i = 1 ; i <= n ; ++ i )
        for ( int j = 1 ; j <= n ; ++ j )
            if ( ! map [i] [j] )  {
                if ( ( i + j ) & 1 )  Lazer.add_edge ( S, id ( i, j ), 1 ) ;
                else Lazer.add_edge ( id ( i, j ), T, 1 ) ;
                if ( ( i + j ) & 1 )  {
                    for ( int k = 0 ; k < 8 ; ++ k )  {
                        int x = i + dx [k], y = j + dy [k] ;
                        if ( x >= 1 && x <= n && y >= 1 && y <= n && ! map [x] [y] )  {
                            Lazer.add_edge ( id ( i, j ), id ( x, y ), oo ) ;
                        }
                    }
                }
            }
    printf ( "%d\n", sum - Lazer.dinic ( S, T ) ) ;
    return 0 ;
}