POJ2446(二分图最大匹配)

时间：2016-09-04 11:34:17 阅读：124 评论：0 收藏：0 [点我收藏+]

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Chessboard

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16924		Accepted: 5284

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).
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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:
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A VALID solution.
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An invalid solution, because the hole of red color is covered with a card.
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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

思路：将两个相邻的grid建边，判断二分图最大匹配*2+k是否等于n*m

#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int MAXN=40;
const int MAXV=1100;
int n,m,k;
int mz[MAXN][MAXN];
int dy[4]={0,1,0,-1};
int dx[4]={1,0,-1,0};
vector<int> arc[MAXV];
int vis[MAXN][MAXN];
int match[MAXV],used[MAXV];
bool dfs(int u)
{
    for(int i=0;i<arc[u].size();i++)
    {
        int to=arc[u][i];
        if(!used[to])
        {
            used[to]=1;
            int w=match[to];
            if(w==-1||dfs(w))
            {
                match[to]=u;
                match[u]=to;
                return true;
            }
        }
    }
    return false;
}
int max_flow()
{
    int ans=0;
    memset(match,-1,sizeof(match));
    int limit=n*m;
    for(int i=1;i<=limit;i++)
    {
        if(match[i]==-1)
        {
            memset(used,0,sizeof(used));
            if(dfs(i))
            {
                ans++;
            }
        }
    }
    return ans;
}
int main()
{
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)    
    {
        if((n*m-k)%2!=0)
        {
            printf("NO\n");
            continue;
        }
        for(int i=0;i<MAXV;i++)    arc[i].clear();
        memset(vis,0,sizeof(vis));
        memset(mz,0,sizeof(mz));
        for(int i=0;i<k;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            mz[y][x]=-1;
        }
        //int edge=0;
        for(int y=1;y<=n;y++)
        {
            for(int x=1;x<=m;x++)
            {
                vis[y][x]=1;
                if(mz[y][x]==-1)    continue;
                for(int i=0;i<4;i++)
                {
                    int ny=y+dy[i];
                    int nx=x+dx[i];
                    if(1<=ny&&ny<=n&&1<=nx&&nx<=m&&mz[ny][nx]!=-1&&!vis[ny][nx])
                    {
        //                edge++;
                        int u=(y-1)*m+x;
                        int v=(ny-1)*m+nx;
                        arc[u].push_back(v);
                        arc[v].push_back(u);    
                    }
                }
            }
        }
        //printf("%d\n",edge);
        int res=max_flow();
        if(res*2+k==n*m)
        {
            printf("YES\n");
        }        
        else
        {
            printf("NO\n");
        }
    }
    return 0;    
}

POJ2446(二分图最大匹配)

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原文地址：http://www.cnblogs.com/program-ccc/p/5838733.html

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