一个 m * n 的棋盘,有的格子存在障碍,求经过所有非障碍格子的哈密顿回路个数

The first line contains the integer numbers N and M (2 ≤ N, M ≤ 12). Each of the next N lines contains M characters, which are the corresponding cells of the rectangle. Character "." (full stop) means a cell, where a segment of the race circuit should be built, and character "*" (asterisk) - a cell, where a gopher hole is located.

You should output the desired number of ways. It is guaranteed, that it does not exceed 2^63-1.

4 4
**..
....
....
....

2

分析：今天把插头dp学了一下，没想到dp还能写这么长......细节什么的也很多，在这里当一个模板吧.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>using namespace std;
typedef long long ll;const int maxn = 30010;
const int pow[13] = {0,2,4,6,8,10,12,14,16,18,20,22,24};
int n,m,now,pre,tx,ty;
char map[20][20];struct node
{int head[maxn],nextt[maxn],tot;ll sum[maxn],sta[maxn];void clear(){memset(head,-1,sizeof(head));tot = 0;}void push(ll x,ll v){ll hashh = x % maxn;for (int i = head[hashh]; i >= 0; i = nextt[i]){if (sta[i] == x){sum[i] += v;return;}}sta[tot] = x;sum[tot] = v;nextt[tot] = head[hashh];head[hashh] = tot++;}
} f[2];int turnleft(ll x,int k)
{return x << pow[k];
}int get(ll x,int k)
{return (x >> pow[k]) & 3;
}ll del(ll x,int i,int j)
{return x & (~(3 << pow[i])) & (~(3 << pow[j]));
}int findr(ll x,int pos)
{int cnt = 1;for (int i = pos + 1; i <= m; i++){int k = get(x,i);if (k == 1)cnt++;else if (k == 2)cnt--;if (!cnt)return i;}
}int findl(ll x,int pos)
{int cnt = 1;for (int i = pos - 1; i >= 0; i--){int k = get(x,i);if (k == 2)cnt++;else if (k == 1)cnt--;if (!cnt)return i;}
}void solve2(int x,int y,int k)
{int p = get(f[pre].sta[k],y - 1);   //右插头int q = get(f[pre].sta[k],y);   //下插头ll staa = del(f[pre].sta[k],y - 1,y);   //将这两个插头删掉以后的状态ll v = f[pre].sum[k];if (!p && !q)  //新建一个连通分量
    {if (map[x][y] == '*'){f[now].push(staa,v);return;}if (x < n && y < n && map[x + 1][y] == '.' && map[x][y + 1] == '.')f[now].push(staa | turnleft(1,y - 1) | turnleft(2,y),v);}else if (!p || !q)  //保持原来的连通分量
    {int temp = p + q;if (x < n && map[x + 1][y] == '.')f[now].push(staa | turnleft(temp,y - 1),v);if (y < m && map[x][y + 1] == '.')f[now].push(staa | turnleft(temp,y),v);}else if (p == 1 && q == 1)    //连接两个联通分量f[now].push(staa ^ turnleft(3,findr(staa,y)),v);  //这里的异或实际上就是把1变成2,2变成1else if (p == 2 && q == 2)f[now].push(staa ^ turnleft(3,findl(staa,y - 1)),v);else if (p == 2 && q == 1)f[now].push(staa,v);else if (x == tx && y == ty)f[now].push(staa,v);
}ll solve()
{f[0].clear();f[0].push(0,1);now = 0,pre = 1;  //滚动数组for (int i = 1; i <= n; i++){pre = now;now ^= 1;f[now].clear();for (int k = 0; k < f[pre].tot; k++)f[now].push(turnleft(f[pre].sta[k],1),f[pre].sum[k]);   //左移一位，因为轮廓线下来的时候会少一个插头for (int j = 1; j <= m; j++){pre = now;now ^= 1;f[now].clear();for (int k = 0; k < f[pre].tot; k++)solve2(i,j,k);  //处理第k个状态
        }}for (int i = 0; i < f[now].tot; i++)if (f[now].sta[i] == 0)  //没有插头了.return f[now].sum[i];return 0;
}int main()
{scanf("%d%d",&n,&m);for(int i = 1; i <= n; i++)scanf("%s",map[i] + 1);for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)if (map[i][j] == '.')tx = i,ty = j; //找右下角的非障碍点if (!tx)puts("0");elseprintf("%lld\n",solve());return 0;
}