poj2446_二分图

POJ 2446  二分图的最大匹配 匈牙利算法

Chessboard

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14350		Accepted: 4471

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

题意：用2*1的木板覆盖给定的棋盘（棋盘中有墙），木板不能重叠，问是否能全部覆盖
思路：奇偶建二分图，把棋盘涂成黑白相间，每个黑格和4个白格相邻，每个白格和四个黑格相邻，黑格移入U,白格移入V，求最大匹配，由于每个匹配对应两个结点，若最大匹配*2等于总结点数说明成功覆盖

/* poj2446 141ms  */

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<algorithm>

#include<vector>



using namespace std;



const int maxn=1200;



int N,M,K;

int uN,vN;

vector<int> G[maxn];

int link[maxn];

bool vis[maxn];

int ch[maxn][maxn];

int dx[]={-1,1,0,0};

int dy[]={0,0,-1,1};



bool dfs(int u)

{

    for(int i=0;i<G[u].size();i++){

        int v=G[u][i];

        if(!vis[v]){

            vis[v]=1;

            if(link[v]==-1||dfs(link[v])){

                link[v]=u;

                return true;

            }

        }

    }

    return false;

}



int hungary()

{

    int res=0;

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

    for(int u=1;u<=uN;u++){

        memset(vis,0,sizeof(vis));

        if(dfs(u)) res++;

    }

    return res;

}



int main()

{

    cin>>N>>M>>K;

    for(int i=1;i<=N*M;i++) G[i].clear();

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

    uN=vN=0;

    while(K--){

        int i,j;

        scanf("%d%d",&j,&i);

        ch[i][j]=0;

    }

    for(int i=1;i<=N;i++){

        for(int j=1;j<=M;j++){

            if(ch[i][j]){

                if((i+j)%2) ch[i][j]=++uN;

                else ch[i][j]=++vN;

            }

        }

    }

    for(int i=1;i<=N;i++){

        for(int j=1;j<=M;j++){

            if(ch[i][j]&&ch[i][j]!=-1&&(i+j)%2){

                for(int k=0;k<4;k++){

                    int x=i+dx[k],y=j+dy[k];

                    if(ch[x][y]&&ch[x][y]!=-1) G[ch[i][j]].push_back(ch[x][y]);

                }

            }

        }

    }

    int match=hungary();

    if(match*2==uN+vN) cout<<"YES"<<endl;

    else cout<<"NO"<<endl;

    return 0;

}

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