HDU1693_Eat the Trees_插头DP

题目大意

插头DP模板题 可形成多个闭合回路

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define INF 0x3f3f3f3f
#define rep0(i, n) for (int i = 0; i < n; i++)
#define rep1(i, n) for (int i = 1; i <= n; i++)
#define rep_0(i, n) for (int i = n - 1; i >= 0; i--)
#define rep_1(i, n) for (int i = n; i > 0; i--)
#define MAX(x, y) (((x) > (y)) ? (x) : (y))
#define MIN(x, y) (((x) < (y)) ? (x) : (y))
#define mem(x, y) memset(x, y, sizeof(x))
#define MAXN 13

using namespace std;
typedef long long LL;
int n, m, mp[MAXN][MAXN];
LL dp[MAXN][MAXN][1 << MAXN];
LL work()
{
    dp[0][m][0] = 1;    //虚拟状态  (0, 0)位置之前的轮廓线 
    for (int i = 1; i <= n + 1; i++)     //递推至n + 1行(全是障碍格)
    {
        for (int k = 0; k < (1 << m); k++)
            dp[i][0][k << 1] = dp[i - 1][m][k];    
        for (int j = 1; j <= m; j++)
        {

            if (mp[i][j])
            {
                for (int k = 0; k < (1 << (m + 1)); k++)
                {
                    int left = 1 << (j - 1);
                    int up = 1 << j;

                    dp[i][j][k] += dp[i][j - 1][k ^ left ^ up];
                    if ((k & left) && (k & up))
                        continue;
                    if ((k & left) == 0 && (k & up) == 0)
                        continue;
                    dp[i][j][k] += dp[i][j - 1][k];

                }
            }
            else
            {
                for (int k = 0; k < (1 << (m + 1)); k++)
                {
                    int left = 1 << (j - 1);
                    int up = 1 << j;

                    if ((k & left) == 0 && (k & up) == 0)
                        dp[i][j][k] = dp[i][j - 1][k];

                }


            }
        }

    }


    return dp[n + 1][m][0];


}
int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt", "r", stdin);
    #endif // ONLINE_JUDGE
    int t, kase = 0;
    scanf("%d", &t);
    while (t--)
    {
        mem(mp, 0);
        mem(dp, 0);
        scanf("%d %d", &n, &m);

        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= m; j++)
            {
                scanf("%d", &mp[i][j]);

            }



        }

        printf("Case %d: There are %lld ways to eat the trees.\n", ++kase, work());

    }

    return 0;
}