POJ 2446 Chessboard

Chessboard

Time Limit: 2000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 2446
64-bit integer IO format: %lld      Java class name: Main

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.

 

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

 

解题：最大匹配

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 33*33;
bool mp[maxn][maxn],used[maxn],cant[50][50];
int Link[maxn],n,m,k;
bool match(int u){
    for(int i = 0; i < maxn; ++i){
        if(!mp[u][i] || used[i]) continue;
        used[i] = true;
        if(Link[i] == -1 || match(Link[i])){
            Link[i] = u;
            return true;
        }
    }
    return false;
}
int main(){
    int x,y;
    while(~scanf("%d%d%d",&n,&m,&k)){
        memset(cant,false,sizeof cant);
        for(int i = 0; i < k; ++i){
            scanf("%d%d",&y,&x);
            cant[x-1][y-1] = true;
        }
        memset(mp,false,sizeof mp);
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < m; ++j){
                if(((i+j)&1) && !cant[i][j]){
                    if(i > 0 && !cant[i-1][j]) mp[i*m + j][(i-1)*m + j] = true;
                    if(j > 0 && !cant[i][j-1]) mp[i*m + j][i*m + j - 1] = true;
                    if(i + 1 < n && !cant[i+1][j]) mp[i*m + j][(i+1)*m + j] = true;
                    if(j + 1 < m && !cant[i][j+1]) mp[i*m + j][i*m + j + 1] = true;
                }
            }
        }
        int ret = 0;
        memset(Link,-1,sizeof Link);
        for(int i = 0; i <= m*n; ++i){
            memset(used,false,sizeof used);
            if(match(i)) ++ret;
        }
        printf("%s\n",n*m - k == 2*ret?"YES":"NO");
    }
    return 0;
}