E - Chessboard

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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.

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 22 13 3

Sample Output

YES

Hint

A possible solution for the sample input.

 

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                using namespace std;int mp[1230][1230],link[1230],mark[1230],g[35][35],ans,opt[35][35];int n,m,k,temp;bool dfs(int x){ for(int i=1;i<=temp;i++) { if(mark[i]==-1&&mp[x][i]) { mark[i]=1; if(link[i]==-1||dfs(link[i])) { link[i]=x; return true; } } } return false;}int main(){ int x,y; while(scanf("%d%d%d",&n,&m,&k)!=EOF) { temp=0; ans=0; memset(link,-1,sizeof(link)); memset(g,0,sizeof(g)); memset(opt,0,sizeof(opt)); for(int i=1;i<=k;i++) { scanf("%d%d",&y,&x); opt[x][y]=1; } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { if(!opt[i][j]) { g[i][j]=++temp; } } } memset(mp,0,sizeof(mp)); for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { if(g[i][j]!=0) { if(i>1&&g[i-1][j]!=0) mp[g[i][j]][g[i-1][j]]=1; if(i
                
                 1&&g[i][j-1]!=0) mp[g[i][j]][g[i][j-1]]=1; if(j