POJ PKU 2446 Chessboard 解题报告

题目链接：http://acm.pku.edu.cn/JudgeOnline/problem?id=2446

这是一道匹配题，把行数（r）和列数（c）按（r+c）%2分成两组，然后连边，做一次二分图匹配，可以直接用匈牙利算法

匹配数等于两组的元素个数则为YES，否则NO

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
using namespace std;

int py[4][2] = { {1, 0}, {-1, 0}, {0, 1}, {0, -1} };
/**
 * 最大二分图匹配
 * 最大二分图匹配 = 最小点覆盖
 * Base 1
 */
#define MAXM 1024 //行
#define MAXN 1024 //列
bool mat[MAXM][MAXN] = {false};
bool isMatchedR[MAXN] = {false};
int matchedR[MAXN];//调用前初始化: memset(matchedR, 0, sizeof(matchedR));

bool getLeftPath(bool matin[MAXM][MAXN], int pos, const int &n)
{
    int i;
    for(i = 1 ; i <= n ; i ++)
    {
        if(matin[pos][i] && !isMatchedR[i])
        {
            isMatchedR[i] = true;
            if(!matchedR[i] || getLeftPath(matin, matchedR[i], n))
            {
                matchedR[i] = pos;
                return true;
            }
        }
    }

    return false;
}
long GetMaxMacthNumber(bool matin[MAXM][MAXN], const int &m, const int &n)
{
    int i,ans = 0;
    for(i = 1 ; i <= m ; i ++)
    {
        memset(isMatchedR, false, sizeof(isMatchedR));
        if(getLeftPath(matin, i, n))
            ans++;
    }
    return ans;
}

bool mapbs[35][35];
int index[35][35];
int main()
{
    int m, n, k, i, j, r, c, tx, ty;
    scanf("%d %d %d", &n, &m, &k);
    memset(matchedR, 0, sizeof(matchedR));
    memset(mapbs, true, sizeof(mapbs));
    for(i = 0; i < k; i ++)
    {
        scanf("%d %d", &c, &r);
        mapbs[r][c] = false;
    }
    for(i = 0; i < 35; i ++)
        mapbs[0][i] = mapbs[i][0] = mapbs[n + 1][i] = mapbs[i][m + 1] = false;

    r = c = 0;
    for(i = 1; i <= n; i ++)
    {
        for(j = 1; j <= m; j ++)
        {
            if(mapbs[i][j])
            {
                if((i + j) % 2 == 0)
                    index[i][j] = ++ r;
                else
                    index[i][j] = ++ c;
            }
        }
    }
    for(i = 1; i <= n; i ++)
    {
        for(j = 1; j <= m; j ++)
        {
            if(mapbs[i][j] && (i + j) % 2 == 0)
            {
                for(k = 0; k < 4; k ++)
                {
                    tx = i + py[k][0];
                    ty = j + py[k][1];
                    if(mapbs[tx][ty])
                    {
                        mat[index[i][j]][index[tx][ty]] = true;
                    }
                }
            }
        }
    }

    if(r != c || GetMaxMacthNumber(mat, r, c) != r)
        printf("NO\n");
    else
        printf("YES\n");
    return 0;
}