poj2446

二分图匹配，每个点只判断与其周围各点的匹配情况，难点在于对于一个编号求期所在的行列，以及对于给出的坐标求编号。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

#define maxn 70

int n, m, h;
bool map[maxn][maxn];
int xM[maxn * maxn], yM[maxn * maxn];
bool chk[maxn * maxn];
int uN, vN;
int dir[4][2] =
{
{ 1, 0 },
{ 0, 1 },
{ -1, 0 },
{ 0, -1 } };

int getx(int a)
{
    if (m & 1)
        return (a * 2 + 1) / m;
    return a / (m / 2);
}

int gety(int a)
{
    if (m & 1)
        return (a * 2 + 1) % m;
    int x = getx(a);
    return 1 - x % 2 + (a * 2 - x * m);
}

int hash(int x, int y)
{
    if (x < 0 || y < 0 || x >= n || y >= m)
        return -1;
    if (map[x][y])
        return -1;
    if (m & 1)
        return (x * m + y) / 2;
    return x * (m / 2) + y / 2;
}

bool SearchPath(int u)
{
    int x = getx(u);
    int y = gety(u);
    if (map[x][y])
        return false;
    for (int i = 0; i < 4; i++)
    {
        int v = hash(x + dir[i][0], y + dir[i][1]);
        if (v != -1 && !chk[v])
        {
            chk[v] = true;
            if (yM[v] == -1 || SearchPath(yM[v]))
            {
                yM[v] = u;
                xM[u] = v;
                return true;
            }
        }
    }
    return false;
}

int MaxMatch()
{
    int u, ret = 0;

    memset(xM, -1, sizeof(xM));
    memset(yM, -1, sizeof(yM));
    for (u = 0; u < uN; u++)
        if (xM[u] == -1)
        {
            memset(chk, 0, sizeof(chk));
            if (SearchPath(u))
                ret++;
        }
    return ret;
}

int main()
{
    //freopen("t.txt", "r", stdin);
    scanf("%d%d%d", &n, &m, &h);
    memset(map, 0, sizeof(map));
    uN = m * n / 2;
    vN = m * n - uN;
    for (int i = 0; i < h; i++)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        a--;
        b--;
        map[b][a] = true;
    }
    if ((m * n - h) & 1)
    {
        printf("NO\n");
        return 0;
    }
    int ans = MaxMatch();
    if (ans < (m *n - h)/2)
        printf("NO\n");
    else
        printf("YES\n");
    return 0;
}