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;
}