Time Limit: 2000MS | Memory Limit: 65536KB | 64bit IO Format: %lld & %llu |
Description
Alice and Bob often play games on chessboard. One day, Alice draws a board with size M * N. She wants Bob to use a lot of cards with size 1 * 2 to cover the board. However, she thinks it too easy to bob, so she makes some holes on the board (as shown in the figure below).

We call a grid, which doesn’t contain a hole, a normal grid. Bob has to follow the rules below:
1. Any normal grid should be covered with exactly one card.
2. One card should cover exactly 2 normal adjacent grids.Some examples are given in the figures below:
A VALID solution.

An invalid solution, because the hole of red color is covered with a card.

An invalid solution, because there exists a grid, which is not covered.
Your task is to help Bob to decide whether or not the chessboard can be covered according to the rules above.

We call a grid, which doesn’t contain a hole, a normal grid. Bob has to follow the rules below:
1. Any normal grid should be covered with exactly one card.
2. One card should cover exactly 2 normal adjacent grids.Some examples are given in the figures below:

A VALID solution.

An invalid solution, because the hole of red color is covered with a card.

An invalid solution, because there exists a grid, which is not covered.
Your task is to help Bob to decide whether or not the chessboard can be covered according to the rules above.
Input
There are 3 integers in the first line: m, n, k (0 < m, n <= 32, 0 <= K < m * n), the number of rows, column and holes. In the next k lines, there is a pair of integers (x, y) in each line, which represents a hole in the y-th row, the x-th column.
Output
If the board can be covered, output “YES”. Otherwise, output “NO”.
Sample Input
4 3 2 2 1 3 3
Sample Output
YES
Source
POJ Monthly,charlescpp
二分图匹配的题,建模真的很难啊,看了题解才做出来的。将整个棋盘看成类似国际象棋棋盘那样黑白相间(忽略hole),然后在相邻点之间加边,便形成了一个二分图,求最大匹配并判断匹配数是否与点的个数相同,即该匹配是否为完全匹配,输出答案即可。
AC代码:
#include <iostream> #include <cstring> using namespace std; const int maxn = 35; int g[maxn][maxn]; bool hole[maxn][maxn]; int id[maxn][maxn]; int cnt; int linker[maxn * maxn]; bool vis[maxn * maxn]; int m, n, k; const int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0}; void input() { cin >> m >> n >> k; int x, y; for (int i = 0; i < k; i++) { cin >> y >> x; hole[x][y] = 1; } } bool dfs(int u) { for (int v = 1; v <= cnt; v++) if (g[u][v] && !vis[v]) { vis[v] = 1; if (linker[v] == -1 || dfs(linker[v])) { linker[v] = u; return 1; } } return 0; } int hungary() { int res = 0; memset(linker, -1, sizeof(linker)); for (int u = 1; u <= cnt; u++) { memset(vis, 0, sizeof(vis)); if (dfs(u)) res++; } return res; } void solve() { for (int i = 1; i <= m; i++) for (int j = 1; j <= n; j++) if (!hole[i][j]) { id[i][j] = ++cnt; for (int d = 0; d < 4; d++) if (id[i + dx[d]][j + dy[d]]) g[id[i][j]][id[i + dx[d]][j + dy[d]]] = 1; } if (hungary() == cnt) cout << "YES\n"; else cout << "NO\n"; } int main() { input(); solve(); return 0; }
0 条评论