Chessboard

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 12800		Accepted: 4000

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.

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.

If the board can be covered, output "YES". Otherwise, output "NO".

4 3 2

2 1

3 3

YES

 
A possible solution for the sample input.

POJ Monthly,charlescpp

 //224K    125MS    C++    1731B    2014-06-10 12:44:41

 #include<iostream>

 #include<vector>

 #define N 1050

 using namespace std;

 vector<int>V[N];

 int match[N];

 int vis[N];

 int g[][];

 int dfs(int u)

 {

     for(int i=;i<V[u].size();i++){

         int v=V[u][i];

         if(!vis[v]){

             vis[v]=;

             if(match[v]==- ||  dfs(match[v])){

                 match[v]=u;

                 return ;

             }

         }

     }

     return ;

 }

 int hungary(int n)

 {

     int ret=;

     memset(match,-,sizeof(match));

     for(int i=;i<=n;i++){

         memset(vis,,sizeof(vis));

         ret+=dfs(i);

     }

     return ret;

 }

 int main(void)

 {

     int n,m,k,x,y;

     while(scanf("%d%d%d",&n,&m,&k)!=EOF)

     {

         memset(g,,sizeof(g));

         for(int i=;i<N;i++) V[i].clear();

         for(int i=;i<k;i++){

             scanf("%d%d",&y,&x);

             g[x-][y]=;

         }

         int map[N]={},pos=;

         for(int i=;i<n;i++)

             for(int j=;j<=m;j++)

                 if(!g[i][j]){

                     if(!map[i*m+j]) map[i*m+j]=++pos;

                     int u=map[i*m+j];

                     if(j<m && !g[i][j+]){

                         if(!map[i*m+j+]) map[i*m+j+]=++pos;

                         V[u].push_back(map[i*m+j+]);

                         V[map[i*m+j+]].push_back(u);

                     }

                     if(i<n- &&  !g[i+][j]){

                         if(!map[(i+)*m+j]) map[(i+)*m+j]=++pos;

                         V[u].push_back(map[(i+)*m+j]);

                         V[map[(i+)*m+j]].push_back(u);

                     }

                 }

         //printf("%d\n",pos);

         if(hungary(pos)==pos) puts("YES");

         else puts("NO");

     }

     return ;

 }