Number of Amidakuji - 牛客

\(dp[i][j]:\) 表示从\((0,0)\)走到\((i,j)\)的方案数
\[ dp[i][j]=\begin{cases} dp[i][j] + dp[i - 1][j - 1] & (从左边过来) \dp[i][j] + dp[i - 1][j + 1] & (从右边过来) \dp[i][j] + dp[i - 1][j] \end{cases} \]

代码

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

const int mod = 1000000007;

int h, w, k;
int dp[200][10];

int main()
{
    cin >> h >> w >> k;

    dp[0][0] = 1;

    for (int i = 1; i <= h; ++i) {
        for(int k = 0; k < (1 << (w - 1)); ++k) {
            
            bool flag = true;
            for (int j = 0; j < (w - 2); ++j) if ((k & (1 << j)) && (k & (1 << (j + 1)))) flag = false;
            if (!flag) continue;

            for (int j = 0; j < w; ++j) {
                if (j >= 1 && (k & (1 << (j - 1)))) {
                    dp[i][j] += dp[i - 1][j - 1];
                    dp[i][j] %= mod;
                }
                else if (j + 1 < w && (k & (1 << j))) {
                    dp[i][j] += dp[i - 1][j + 1];
                    dp[i][j] %= mod;
                }
                else {
                    dp[i][j] += dp[i - 1][j];
                    dp[i][j] %= mod;
                }
            } 
        }
    }

    printf("%d\n", dp[h][k - 1]);

    return 0;
}

Number of Amidakuji - 牛客

原文：https://www.cnblogs.com/zgglj-com/p/10196485.html