POJ 2446 最小点覆盖

Chessboard

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14787		Accepted: 4607

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

 

 

题目意思：

一个n*m的棋盘，现用1*2的木板覆盖棋盘，其中有黑色圆的棋盘单位不能被覆盖。问能否把棋盘中可以覆盖的单位全部覆盖（不能重复覆盖）。

 

思路：

相邻棋盘单位奇偶性不同建图，判断最大匹配*2是否等于可以覆盖点数即可。

 

 

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <cmath>
#include <set>
using namespace std;

#define N 35

int max(int x,int y){return x>y?x:y;}
int min(int x,int y){return x<y?x:y;}
int abs(int x,int y){return x<0?-x:x;}

int n, m;
vector<int>ve[N*N];
int from[N*N];
bool visited[N*N];


int march(int u){
    int i, v;
    for(i=0;i<ve[u].size();i++){
        v=ve[u][i];
        if(!visited[v]){
            visited[v]=true;
            if(from[v]==-1||march(from[v])){
                from[v]=u;
                return 1;
            }
        }
    }
    return 0;
}
int map[N][N];
main()
{
    int i, j, k;
    
    int x, y;
    while(scanf("%d %d %d",&n,&m,&k)==3){
        memset(map,-1,sizeof(map));
        int maxh=0;
        for(i=0;i<n;i++){
            for(j=0;j<m;j++){
                if(j==0){
                    if(i==0) map[i][j]=0;
                    else {
                        if(m&1) map[i][j]=map[i-1][m-1]+1;
                        else map[i][j]=map[i-1][m-1]+2;
                    }
                }
                else map[i][j]=map[i][j-1]+1;
                maxh=max(maxh,map[i][j]);
            //    printf("%d ",map[i][j]);
            }
            //cout<<endl;
        }
        for(i=0;i<k;i++){
            scanf("%d %d",&x,&y);
            map[y-1][x-1]=-1;
        } 
        int nn=0;
        for(i=0;i<n;i++){
            for(j=0;j<m;j++){
                if(map[i][j]==-1) nn++;
            }
        }
        
        for(i=0;i<=maxh;i++) ve[i].clear();
        for(i=0;i<n;i++){
            for(j=0;j<m;j++){
                if(map[i][j]!=-1&&(map[i][j]&1)){
                    if(i>0&&map[i-1][j]!=-1) ve[map[i][j]].push_back(map[i-1][j]);
                    if(i<n-1&&map[i+1][j]!=-1) ve[map[i][j]].push_back(map[i+1][j]);
                    if(j>0&&map[i][j-1]!=-1) ve[map[i][j]].push_back(map[i][j-1]);
                    if(j<m-1&&map[i][j+1]!=-1) ve[map[i][j]].push_back(map[i][j+1]);
                }
            }
        }
        int num=0;
        memset(from,-1,sizeof(from));
        for(i=0;i<=maxh;i++){
            memset(visited,false,sizeof(visited));
            if((i&1)&&march(i)) num++;
        }
        if(num*2==n*m-nn) printf("YES
");
        else printf("NO
"); 
    }
}