很容易想到是最大匹配：行和列看做点，每一个可放置的位置看做边，先求一遍最大匹配，然后枚举删除每一条边，如果删除了某一条边发现求出来的最大匹配比之前少，则这条边对应的格点是”重要点“。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 6 const int N = 101; 7 bool mp[N][N]; 8 int mark[N]; 9 bool visit[N]; 10 int n, m, k; 11 12 int dfs( int u ) 13 { 14 for ( int i = 1; i <= m; i++ ) 15 { 16 if ( !visit[i] && mp[u][i] ) 17 { 18 visit[i] = 1; 19 if ( mark[i] == -1 || dfs(mark[i]) ) 20 { 21 mark[i] = u; 22 return 1; 23 } 24 } 25 } 26 return 0; 27 } 28 29 inline int hungary() 30 { 31 int res = 0; 32 memset( mark, -1, sizeof(mark) ); 33 for ( int i = 1; i <= n; i++ ) 34 { 35 memset( visit, 0, sizeof(visit) ); 36 res += dfs(i); 37 } 38 return res; 39 } 40 41 int main() 42 { 43 int _case = 1; 44 while ( scanf("%d%d%d", &n, &m, &k) != EOF ) 45 { 46 memset( mp, 0, sizeof(mp) ); 47 while ( k-- ) 48 { 49 int x, y; 50 scanf("%d%d", &x, &y); 51 mp[x][y] = 1; 52 } 53 int cnt = 0, ans = hungary(); 54 for ( int i = 1; i <= n; i++ ) 55 { 56 for ( int j = 1; j <= m; j++ ) 57 { 58 if ( !mp[i][j] ) continue; 59 mp[i][j] = 0; 60 if ( hungary() < ans ) cnt++; 61 mp[i][j] = 1; 62 } 63 } 64 printf("Board %d have %d important blanks for %d chessmen. ", _case++, cnt, ans); 65 } 66 return 0; 67 }