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;
}