POJ 2446 Chessboard（匈牙利算法）

Chessboard

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 19211 Accepted: 6063

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.

Source

POJ Monthly,charlescpp

【思路】

将棋盘像国际象棋棋盘那样分割为黑白块，那么用纸片覆盖时就可以视作匹配没有洞的黑块和白块，用匈牙利算法求二分图的最大匹配即可。具体可以这么建图，二分图的左右点集都是黑白块1到N*M，若黑块可能与周边白块相匹配，则从左点集对应的黑块与右点集对应的白块连一条线。跑一遍匈牙利算法后，看二倍最大匹配数是否等于N*M-K就好了。

【代码】

#include<cstdio>#include<cstring>#include<vector>#include<algorithm>using namespace std;const int MAXN=100;const int DIR[4][2]={{0,-1},{-1,0},{0,1},{1,0}};int m,n,k;bool visited[MAXN*MAXN];bool hole[MAXN][MAXN];int xmatch[MAXN*MAXN],ymatch[MAXN*MAXN];vector<int> connect[MAXN];bool dfs(int u){    for(int i=0;i<connect[u].size();i++){        int j=connect[u][i];        if(!visited[j]){            visited[j]=true;            if(ymatch[j]==-1||dfs(ymatch[j])){                xmatch[u]=j;                ymatch[j]=u;                return true;            }        }    }    return false;}int hungary(){    int ans=0;    memset(xmatch,0xff,sizeof(xmatch));    memset(ymatch,0xff,sizeof(ymatch));    for(int i=1;i<=n*m;i++){        memset(visited,false,sizeof(visited));        if(dfs(i))ans++;    }    return ans;}int main(){    memset(hole,false,sizeof(hole));    scanf("%d %d %d",&m,&n,&k);    for(int i=1;i<=k;i++){        int x,y;        scanf("%d %d",&x,&y);        hole[y][x]=true;    }    for(int i=1;i<=m;i++)        for(int j=1;j<=n;j++)            if(!hole[i][j])                for(int k=0;k<4;k++){                    int y=i+DIR[k][0],x=j+DIR[k][1];                    if(1<=y&&y<=m&&1<=x&&x<=n&&!hole[y][x]){                        connect[(i-1)*n+j].push_back((y-1)*n+x);                    }                }    if(hungary()==n*m-k)        printf("YES\n");    else        printf("NO\n");    return 0;}