AcWing 327. 玉米田
原题链接
简单
作者:
chenjiaqiy
,
2023-12-10 15:41:22
,
所有人可见
,
阅读 56
#include<iostream>
#include <vector>
#include<algorithm>
using namespace std;
const int N = 14,M = 1 << 12,mod = 1e8;
int n,m;
int g[N];
vector <int> state;
vector<int> head[M];
int f[N][M];
bool check(int state)
{
for (int i = 0; i < m; i ++ )
{
if((state >> i & 1) && (state >> i + 1 & 1))
return false;
}
return true;
}
int main()
{
cin >> n >> m;
for (int i = 1; i <= n; i ++ )
for (int j = 0; j < m; j ++ )
{
int t;
cin >> t;
g[i] += !t << j; //把地图转化为二进制地图
}
for (int i = 0; i < 1 << m; i ++ ) //枚举一下列的所有状态
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]; //a,b简化两层状态
if((a & b) == 0)
head[i].push_back(j); //i可以转移到j
}
f[0][0] = 1; //一行没有,一个玉米没种方案数为1
for (int i = 1; i <= n + 1; i ++ )
for (int a = 0; a < state.size(); a ++ )
for (int b : head[a]) //a能到的状态
{
if(g[i] & state[a]) continue; //田地状态不满足
f[i][a] = (f[i - 1][b] + f[i][a]) % mod;
}
cout << f[n + 1 ][0] << endl;
}