hihocoder_offer收割编程练习赛52_2亮灯方案

题目链接： https://hihocoder.com/contest/offers52/problem/2

解题思路： 题目中的关键约束条件是n*m<=20，所以总的状态空间是2^20个。从初始状态，依次枚举每个可能的操作，统计方案总数。

#include <bits/stdc++.h>
using namespace std;

char s[21];
bool a[21][21];
int n, m;
map<int, long long> vis;
long long ans;

int judge(int i, int j)
{
    if (i - 1 >= 0 && a[i-1][j] == 1) return 1;
    if (i + 1 < n && a[i+1][j] ==1 )return 1;
    if (j-1 >= 0 && a[i][j-1] == 1) return 1;
    if (j + 1 < m && a[i][j+1] == 1) return 1;
    return 0;
}

int array_to_int(bool a[21][21])
{
    int res = 0;
    for (int i = 0; i < n; ++i)
    {
        for (int j = 0; j < m; ++j)
        {
            res = res*2 + a[i][j];
        }
    }
    return res;
}

long long dfs(bool a[21][21])
{
    int hasha = array_to_int(a);
    if (vis.find(hasha) != vis.end())
    {
        return vis[hasha];
    }
    int flags = 0;
    long long res = 0;
    for (int k = 0; k < n*m; ++k)
    {
        int i = k / m;
        int j = k % m;
        if (a[i][j] == 0 && judge(i, j))
        {
            a[i][j] = 1;
            res += dfs(a);
            a[i][j] = 0;
            flags = 1;
        }
    }
    if (!flags)
        res = 1;
    vis[hasha] = res;
    return res;
}


int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; ++i)
    {
        scanf("%s", s);
        for (int j = 0; j < m; ++j)
        {
            a[i][j] = (s[j]=='1'?1:0);
        }
    }
    long long ans = dfs(a);
    printf("%lld\n", ans);
    return 0;
}