Chessboard------最大独立数

Chessboard

Time Limit(Common/Java):1000MS/3000MS          Memory Limit:65536KByte
Total Submit:12            Accepted:12

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 2
2 1
3 3

 

Sample Output

 

YES

 

Hint

A possible solution for the sample input.

 

Source

HHUACM

 

#include<iostream>#include<vector>using namespace std;vector< vector<int> >adj;int a[33][33]={0},m,n,K,loc=0;int pre[500]={0},flag[1009];void get_a(int i,int j){int t,flag=0;if(a[i][j])return;if(i+1<=m&&a[i+1][j]==0){if(flag==0){flag++;loc++;}t=i*m+j;adj[loc].push_back(t);}if(i-1>=m&&a[i-1][j]==0){if(flag==0){flag++;loc++;}t=(i-2)*m+j;adj[loc].push_back(t);}if(j+1<=m&&a[i][j+1]==0){if(flag==0){flag++;loc++;}t=(i-1)*m+j+1;adj[loc].push_back(t);}if(j-1<=m&&a[i][j-1]==0){if(flag==0){flag++;loc++;}t=(i-1)*m+j-1;adj[loc].push_back(t);}}bool work(int i){int j,t;for(j=0;j<adj[i].size();j++){t=adj[i][j];if(flag[t]==0){if(pre[t]==0||work(pre[t])){pre[t]=i;return true;}}}return false;}int hungary(){int sum=0,i;for(i=1;i<=loc;i++){memset(flag,0,sizeof(flag));if(work(i)){sum++;}}return sum;}int main(){int i,j,p,q;cin>>m>>n>>K;adj.assign(m*n/2+2,vector<int>());for(i=1;i<=K;i++){cin>>q>>p;a[p][q]=1;}for(i=1;i<=m;i++){for(j=1;j<=n;j++){if((i+j)%2==0){get_a(i,j);}}}if(loc==(m*n-K)/2&&hungary()==(m*n-K)/2){cout<<"YES"<<endl;}else{cout<<"NO"<<endl;}return 0;}