POJ2446 二分图最大匹配

问题：POJ2446

分析：

　　采用黑白相间的方法把棋盘分成两个点集，是否可以用1*2的卡片实现全覆盖等价于二分图是否有完全匹配。

AC代码

 //Memory: 172K        Time: 32MS
 #include <iostream>
 #include <cstring>
 #include <cstdio>

 using namespace std;

  *  /  + ;
 ][];
 ];
 int ne[maxn];
 int match[maxn];
 int vis[maxn];
 int m, n, k;
 int x, y;
 int flag;
 int num1, num0;
 ][] = { {, }, {-, }, {, }, {, -}};

 bool findPath(int start)
 {
     ; i < ne[start]; i++){
         int current = edge[start][i];
         if ( vis[current] ) continue;
         vis[current] = ;
         if ( !match[current] || findPath( match[current] ) ){
             match[current] = start;
             return true;
         }
     }
     return false;
 }

 bool solve()
 {
     memset(match, , sizeof(match));
     ;
     ; i <= num0; i++){
         memset(vis, , sizeof(vis));
         if ( findPath(i) )
             ++cnt;
     }
     if (cnt == num0) return true;
     return false;
 }

 void input()
 {
     memset(edge, , sizeof(edge));
     memset(ne, , sizeof(ne));
     memset(maze, , sizeof(maze));
     scanf("%d%d%d", &m, &n, &k);
     ; i < k; i++){
         scanf("%d%d", &x, &y);
         maze[y - ][x - ] =  -;
     }
     num0 = num1 = ;
     ; i < m; i++)
         ; j < n; j++){
              && (i + j) %  == ) {
                 maze[i][j] = ++num0;
             }
              && (i + j) %  == ) {
                 maze[i][j] = ++num1;
             }
         }
     if ( num0 != num1 ) return;
     ; i < m; i++) {
         ; j < n; j++){
              ==  && maze[i][j] != -){
                 int current = maze[i][j];
                 ; k < ; k++){
                     ];
                     ];
                      || nj >= n || nj < )
                         continue;
                     ) edge[current][ne[current]++] = maze[ni][nj];
                 }
             }
         }
     }
 }

 int main()
 {
     input();
     if ( num0 == num1 && solve() ) printf("YES\n");
     else printf("NO\n");
     ;
 }