| 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
Hint
A possible solution for the sample input.
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;
}
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