思路

因为每一行只能有一个車,每一列也只能有一个車.因此把每一行当做一个节点,每一列当做一个节点,若行列能放棋子,行节点向列节点连边,这样就是一个二分图最大匹配问题.跑一遍匈牙利算法即可.

代码

#include<bits/stdc++.h>
using namespace std;
#define MAXN 205
#define i64 long long
#define fp( i, b, e ) for ( int i(b), I(e); i <= I; ++i )
#define fd( i, b, e ) for ( int i(b), I(e); i >= I; --i )
#define go( i, b ) for ( int i(b), v(to[i]); i; v = to[i = nxt[i]] )
template<typename T> inline void cmax( T &x, T y ){ x < y ? x = y : x; }
template<typename T> inline void cmin( T &x, T y ){ y < x ? x = y : x; }
#define getchar() ( p1 == p2 && ( p1 = bf, p2 = bf + fread( bf, 1, 1 << 21, stdin ), p1 == p2 ) ? EOF : *p1++ )
char bf[1 << 21], *p1(bf), *p2(bf);
template<typename T>
inline void read( T &x ){ char t(getchar()), flg(0); x = 0;
    for ( ; !isdigit(t); t = getchar() ) flg = t == '-';
    for ( ; isdigit(t); t = getchar() ) x = x * 10 + ( t & 15 );
    flg ? x = -x : x;
}

clock_t t_bg, t_ed;
int N, M, T, mc[MAXN];
bool v[MAXN][MAXN], vis[MAXN];
vector<int> e[MAXN];

bool DFS( int u ){
    for( int v : e[u] ) if ( !vis[v] ){
        vis[v] = 1; if ( !mc[v] || DFS(mc[v]) ) return mc[v] = u, 1;
    } return 0;
}

int main(){
    t_bg = clock();
    read(N), read(M), read(T);
    while( T-- ){ int x, y; read(x), read(y), v[x][y] = 1; }
    fp( i, 1, N ) fp( j, 1, M ) if ( !v[i][j] ) e[i].push_back(j);
    int ans(0); fp( i, 1, N ) memset( vis, 0, sizeof vis ), ans += DFS(i);
    printf( "%d\n", ans );
    t_ed = clock();
    fprintf( stderr, "\n========info========\ntime : %.3f\n====================\n", (double)( t_ed - t_bg ) / CLOCKS_PER_SEC );
    return 0;
}