【ACWing】327. Corn Field

Subject address:

https://www.acwing.com/problem/content/329/

Farmer John’s land by $M\times It\; is$ composed of$N$ small squares, and now he wants to grow corn in the land. It is a pity that part of the land is sterile and cannot be planted. Moreover, adjacent land cannot be planted with corn at the same time, which means that there will be no common edge between all the squares where corn is planted. Now given the size of the land, please find out how many planting methods there are. Planting nothing on the land is a method.

Input format:
No. $1$ line contains two integers$M$ and$N$ . No.$2,...,M+1$ line: each line contains$N$ integers$0$ or$1$ , used to describe the condition of the entire land,$1$ means the land is fertile,$0$ means that the land is sterile.

Output format:
output total planting method to $10^8$ The value after the modulus.

Data range:
$1\le M,N\le 12$

The idea is dynamic programming, which can enumerate by row, and store the corn planting status of each row in a state compression method. For example, we can use the binary digits of an integer to enumerate whether corn is planted at each position, $1$ means species,$0$ means no species. In addition, if you also use an array$A-$ like$i$ number of binary bits to storeiiThe land condition of row$i$ ,$1$ means it cannot be planted,$0$ means energy, then theiiThe valid states of row$i$ are those that are related to$A\left[i\right]$ does the AND operation equals$0$ and not with the$i-1$ line of contradictory state. Let$F\left[I\right]\left[S\right]$ Shi firstiiThe planting status of row$i$ is$In$ the case of s, how many kinds of planting schemes are there. Then according to the$i-According$ to the planting status of$1$ row, there are:$$f[i][s]=\sum _{(t\to s)\wedge (s\&A[i]=0)}f[i-1]\left[t\right]$$ Initial conditions$f[0][0]=1$ . code show as below:

#include <iostream>
#include <vector>
using namespace std;

const int N = 15, M = 1 << 12, mod = 1e8;
int n, m;
int a[N], f[N][M];
// state存每行的合法状态（即没有两个连续的1的状态），
// head存state[i]能转移到的合法状态（即没有矛盾的状态）
vector<int> state, head[M];

bool check(int st) {
    
    
    for (int i = 0; i < n - 1; i++)
        if ((st >> i & 1) && (st >> i + 1 & 1))
            return false;
    return true;
}

int main() {
    
    
    cin >> m >> n;
    for (int i = 1; i <= m; i++)
        for (int j = 0; j < n; j++) {
    
    
            int s;
            cin >> s;
            a[i] += !s << j;
        }

    for (int i = 0; i < 1 << n; i++)
        if (check(i))
            state.push_back(i);

    for (int i = 0; i < state.size(); i++)
        for (int j = 0; j < state.size(); j++) {
    
    
            int a = state[i], b = state[j];
            if (!(a & b)) head[i].push_back(j);
        }

    f[0][0] = 1;
    for (int i = 1; i <= m; i++)
        for (int cur = 0; cur < state.size(); cur++)
            for (int prev : head[cur])
                if (!(state[cur] & a[i]))
                    f[i][cur] = (f[i][cur] + f[i - 1][prev]) % mod;
    
    int res = 0;
    for (int i = 0; i < state.size(); i++) res = (res + f[m][i]) % mod;

    cout << res << endl;

    return 0;
}

Time complexity $O(m{2}^{2n})$, it’s actually not so high, because the legal status of the line is much smaller than$2^n$ , space$O(m{2}^n)$。