POJ 2446 Chessboard （二分匹配）

Chessboard

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10743		Accepted: 3338

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

 

 

 

简单的而二分图的模板题。

建图方法好像不唯一。

我是把每个空格子当成一个点。

然后对于每个格子如果可以往上、下、左、右去找可以建立的边。

然后求最大匹配，如果最大匹配数刚好等于空格子数，输出YES，否则NO。

#include<stdio.h>

#include<string.h>

#include<algorithm>

#include<iostream>

using namespace std;





/* **************************************************************************

//二分图匹配（匈牙利算法的DFS实现）

//初始化：g[][]两边顶点的划分情况

//建立g[i][j]表示i->j的有向边就可以了，是左边向右边的匹配

//g没有边相连则初始化为0

//uN是匹配左边的顶点数，vN是匹配右边的顶点数

//调用：res=hungary();输出最大匹配数

//优点：适用于稠密图，DFS找增广路，实现简洁易于理解

//时间复杂度:O(VE)

//***************************************************************************/

//顶点编号从0开始的

const int MAXN=1610;

int uN,vN;//u,v数目

int g[MAXN][MAXN];

int linker[MAXN];

bool used[MAXN];

bool dfs(int u)//从左边开始找增广路径

{

    int v;

    for(v=0;v<vN;v++)//这个顶点编号从0开始，若要从1开始需要修改

      if(g[u][v]&&!used[v])

      {

          used[v]=true;

          if(linker[v]==-1||dfs(linker[v]))

          {//找增广路，反向

              linker[v]=u;

              return true;

          }

      }

    return false;//这个不要忘了，经常忘记这句

}

int hungary()

{

    int res=0;

    int u;

    memset(linker,-1,sizeof(linker));

    for(u=0;u<uN;u++)

    {

        memset(used,0,sizeof(used));

        if(dfs(u)) res++;

    }

    return res;

}



bool graph[50][50];

int num[50][50];



int main()

{

    int n,m,k;

    int x,y;

    while(scanf("%d%d%d",&n,&m,&k)==3)

    {

        memset(g,0,sizeof(g));

        memset(graph,false,sizeof(graph));

        while(k--)

        {

            scanf("%d%d",&x,&y);

            x--;

            y--;

            graph[y][x]=true;

        }

        int tol=0;

        for(int i=0;i<n;i++)

          for(int j=0;j<m;j++)

          {

              if(graph[i][j]==false)

                num[i][j]=tol++;

          }

        uN=vN=tol;

        for(int i=0;i<n;i++)

           for(int j=0;j<m;j++)

              if(!graph[i][j])

              {

                  int u=num[i][j];

                  if(i>0 && !graph[i-1][j])g[u][num[i-1][j]]=1;

                  if(j>0 && !graph[i][j-1])g[u][num[i][j-1]]=1;

                  if(i<n-1 && !graph[i+1][j])g[u][num[i+1][j]]=1;

                  if(j<m-1 && !graph[i][j+1])g[u][num[i][j+1]]=1;

              }

        if(hungary()==tol)printf("YES\n");

        else printf("NO\n");



    }

    return 0;

}