递增三元组

此题考查暴力，二分

此题未 AC 用了两种方法解题

1. dfs
2. binarySearch

1. dfs

   package lanqiao;
   
   import java.util.Scanner;
   
   public class N172 {
       static int[][] m;
       static int n;
       static long res = 0;
   
       public static void main(String[] args) {
           Scanner scanner = new Scanner(System.in);
           n = scanner.nextInt();
           m = new int[4][n + 1];
           for (int i = 1; i <= 3; i++) {
               for (int j = 1; j <= n; j++) {
                   m[i][j] = scanner.nextInt();
               }
           }
           for (int i = 1; i <= n; i++)
               dfs(1, i);
           System.out.println(res);
       }
   
       private static void dfs(int a, int b) {
           if (a == 3) {
               res++;
               return;
           }
           for (int i = 1; i <= n; i++) {
               if (a + 1 <= 3)
                   if (m[a + 1][i] > m[a][b]) {
                       dfs(a + 1, i);
                   }
           }
       }
   }
   

   2. binarySearch

   package test;
   
   import java.util.Arrays;
   import java.util.Scanner;
   
   public class N172 {
   	static int n;
   	static int[] a;
   	static int[] b;
   	static int[] c;
   
   	public static void main(String[] args) {
   		Scanner scanner = new Scanner(System.in);
   		n = scanner.nextInt();
   		a = new int[n + 1];
   		b = new int[n + 1];
   		c = new int[n + 1];
   		for (int i = 1; i <= n; i++)
   			a[i] = scanner.nextInt();
   		for (int i = 1; i <= n; i++)
   			b[i] = scanner.nextInt();
   		for (int i = 1; i <= n; i++)
   			c[i] = scanner.nextInt();
   		Arrays.sort(a);
   		Arrays.sort(c);
   		int res = 0;
   		for (int i = 1; i <= n; i++) {
   			int x1 = search2(1, n, b[i], a);
   			int x2 = search(1, n, b[i], c);
   			res += x1 * (n - x2 + 1);
   		}
   		System.out.println(res);
   	}
   
   	/**
   	 * 找到第一个大于 x 的数的下标
   	 * 
   	 * @param l
   	 * @param r
   	 * @param x
   	 * @param a
   	 * @return
   	 */
   	static int search(int l, int r, int x, int[] a) {
   		while (l <= r) {
   			int mid = (l + r) / 2;
   			if (x < a[mid]) {
   				r = mid - 1;
   			} else {
   				l = mid + 1;
   			}
   		}
   		return l;
   	}
   
   	/**
   	 * 找到第一个小于 x 的下标
   	 * 
   	 * @param l
   	 * @param r
   	 * @param x
   	 * @param a
   	 * @return
   	 */
   	static int search2(int l, int r, int x, int[] a) {
   		while (l <= r) {
   			int mid = (l + r) / 2;
   			if (x <= a[mid]) {
   				r = mid - 1;
   			} else {
   				l = mid + 1;
   			}
   		}
   		return r;
   	}
   }
   

   说明：

   1. dfs 超时
   2. bs 有两个测试点答案错误，有大佬知道哪里有错误还望指正，感谢各位！