题解 | #二叉树#

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;

// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
    public static void main(String[] args) throws IOException  {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StreamTokenizer in = new StreamTokenizer(br);
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        while (in.nextToken() != StreamTokenizer.TT_EOF) {
            int n = (int) in.nval;
            in.nextToken();
            int m = (int) in.nval;
            // long[][] dp = new long[n + 1][m + 1];
            // for (int i = 0; i <= n; i++) {
            //     for (int j = 0; j <= m; j++) {
            //         dp[i][j] = -1;
            //     }
            // }
            out.println(compute3(n, m));
        }
        out.flush();
        out.close();
        br.close();
    }

    // 记忆化搜索动态规划
    private static long compute(int n, int m, long[][] dp) {
        if (n == 0) {
            return 1;
        }
        if (m == 0) {
            return 0;
        }
        if (dp[n][m] != -1) {
            return dp[n][m];
        }

        long ans = 0;
        for (int i = 0; i < n; i++) {
            ans = (ans + compute(i, m - 1, dp) * compute(n - i - 1, m - 1,
                    dp)) % 1000000007;
        }
        dp[n][m] = ans;
        return ans;
    }

    // 严格位置依赖的动态规划
    private static long compute2(int n, int m, long[][] dp) {
        for (int j = 0; j <= n; j++) {
            dp[j][0] = 0;
        }
        for (int i = 0; i <= m; i++) {
            dp[0][i] = 1;
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                dp[i][j] = 0;
                for (int k = 0; k < i; k++) {
                    dp[i][j] = (dp[i][j] + dp[k][j - 1] * dp[i - k - 1][j - 1] % 1000000007) %
                               1000000007;
                }
            }
        }
        return dp[n][m];
    }

    //动态规划 + 空间压缩
    private static long compute3(int n, int m) {
        long[] dp = new long[n + 1];
        dp[0] = 1;
        for (int j = 1; j <= m; j++) {

            for (int i = n; i >= 1; i--) {
                dp[i] = 0;
                for (int k = 0; k < i; k++) {
                    dp[i] = (dp[i] + dp[k] * dp[i - k - 1] % 1000000007) % 1000000007;
                }
            }
        }
        return dp[n];
    }
}