poj 2446 Chessboard (二分匹配)

Chessboard

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

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

POJ Monthly,charlescpp

 

 

和 hdu 1507类似，构无向图然后判断匹配数是否等于合法的格数。

心算32*32错了= = RE了两次，开始以为32*32是90+，第二次以为是900+，笔算后才知道是1024..

//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[35][35];
int dfs(int u)
{
    for(int i=0;i<V[u].size();i++){
        int v=V[u][i];
        if(!vis[v]){
            vis[v]=1;
            if(match[v]==-1 ||  dfs(match[v])){
                match[v]=u;
                return 1;
            }
        }
    }
    return 0;
}
int hungary(int n)
{
    int ret=0;
    memset(match,-1,sizeof(match));
    for(int i=1;i<=n;i++){
        memset(vis,0,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,0,sizeof(g));
        for(int i=0;i<N;i++) V[i].clear();
        for(int i=0;i<k;i++){
            scanf("%d%d",&y,&x);
            g[x-1][y]=1;
        }
        int map[N]={0},pos=0;
        for(int i=0;i<n;i++)
            for(int j=1;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+1]){
                        if(!map[i*m+j+1]) map[i*m+j+1]=++pos;
                        V[u].push_back(map[i*m+j+1]);
                        V[map[i*m+j+1]].push_back(u);
                    }
                    if(i<n-1 &&  !g[i+1][j]){
                        if(!map[(i+1)*m+j]) map[(i+1)*m+j]=++pos;
                        V[u].push_back(map[(i+1)*m+j]);
                        V[map[(i+1)*m+j]].push_back(u);
                    }
                }
        //printf("%d\n",pos);
        if(hungary(pos)==pos) puts("YES");
        else puts("NO");
    }
    return 0;
}