HDU 3360 National Treasures 奇偶匹配的最低点覆盖

意甲冠军：假设a[i][j] ！= -1 把他转成二进制最多有12位代表题目那张图的12个位置假设相应位是1 说明在那里放一个守卫能够看住a[i][j]位置上的这个东西

思路：明显死最小点覆盖奇偶匹配建图

#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int maxn = 55;
int vis[maxn*maxn];
int y[maxn*maxn];
vector <int> G[maxn*maxn];
int n, m;
int a[maxn][maxn];
int dir[12][2] = {-1, -2, -2, -1, -2, 1, -1, 2, 1, 2, 2, 1, 2, -1, 1, -2, -1, 0, 0, 1, 1, 0, 0, -1};
bool dfs(int u)
{
    for(int i = 0; i < G[u].size(); i++)
    {
        int v = G[u][i];
        if(vis[v])
            continue;
        vis[v] = true;
        if(y[v] == -1 || dfs(y[v]))
        {
            y[v] = u;
            return true;
        }
    }
    return false;
}
int match()
{
    int ans = 0;
    memset(y, -1, sizeof(y));
    for(int i = 0; i < n; i++)
    {
    	for(int j = 0; j < m; j++)
        {
        	if((i+j)%2)
			{
				memset(vis, 0, sizeof(vis));
        		if(dfs(i*m+j))
        		    ans++;
			}
        }
    }
    return ans;
}

int main()
{
	int cas = 1;
	while(scanf("%d %d", &n, &m) && (n||m))
	{
		for(int i = 0; i < n; i++)
			for(int j = 0; j < m; j++)
				scanf("%d", &a[i][j]);
		for(int i = 0; i < n*m; i++)
			G[i].clear();
		for(int i = 0; i < n; i++)
		{
			for(int j = 0; j < m; j++)
			{
				int x = a[i][j];
				if(x == -1)
					continue;
				for(int k = 0; k < 12; k++, x >>= 1)
				{
					if(!x)
						break;
					if(!(x&1))
						continue;
					
					int xx = i + dir[k][0];
					int yy = j + dir[k][1];
					if(xx < 0 || xx >= n || yy < 0 || yy >= m)
						continue;
					if(a[xx][yy] == -1)
						continue;
					if((i+j)%2)
						G[i*m+j].push_back(xx*m+yy);
					else
						G[xx*m+yy].push_back(i*m+j);
				}
			}
		}
		printf("%d. %d
", cas++, match());
	}
    return 0;
}