poj 2446 Chessboard(经典二分匹配)

poj 2446 Chessboard(经典二分匹配)

题目：​​class="data-table" data-id="t7a7e9d1-TANLPiFA" data-transient-attributes="class" data-width="1162px" style="width: 100%; outline: none; border-collapse: collapse;">

Time Limit: 2000MS

Memory Limit: 65536K

Total Submissions: 15250

Accepted: 4751

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 1 3 3

Sample Output

YES

Hint

A possible solution for the sample input.

分析：一眼能看出这是二分匹配吗？所以我觉得这很经典（当然我也是听了别人的思路才懂的）。一张卡片是1*2的规模，所以一个格子就有上下左右四种可能的相邻的格子与他组合成卡片（没有hole），这就是匹配的雏形，于是用邻接表把所有格子的各种匹配情况都表示出来，最后用匈牙利最大匹配求出结果再与总的格子数，holes数做关系比较就能得到最终的判断。

#include #include #include using namespace std;const int N=2000;bool bmap[N][N],bmask[N],tag[35][35];int nx,ny,cx[N],cy[N];int find(int u){ for(int j=1;j<=ny;j++){ if(bmap[u][j]&&!bmask[j]){ bmask[j]=1; if(cy[j]==-1||find(cy[j])){ cy[j]=u; cx[u]=j; return 1; } } } return 0;}int maxmatch(){ int res=0; memset(cx,-1,sizeof(cx)); memset(cy,-1,sizeof(cy)); for(int i=1;i<=nx;i++){ if(cx[i]==-1){ for(int j=1;j<=ny;j++) bmask[j]=0; res+=find(i); } } return res;}int dir[4][2]={{0,-1},{0,1},{-1,0},{1,0}};int main(){ //freopen("cin.txt","r",stdin); int m,n,k; while(cin>>m>>n>>k){ int a,b; nx=ny=n*m; memset(tag,0,sizeof(tag)); memset(bmap,0,sizeof(bmap)); for(int i=0;i=1&&y<=n&&y>=1&&tag[x][y]==0){ a=(i-1)*n+j; b=(x-1)*n+y; bmap[a][b]=1; } } } } int res=maxmatch(); //cout<

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