https://vjudge.net/contest/317762#problem/L
Given a set of digits S, and an integer n, you have to find how many n-digit integers are there, which contain digits that belong to S and the difference between any two adjacent digits is not more than two.
Input
Input starts with an integer T (≤ 300), denoting the number of test cases.
Each case contains two integers, m (1 ≤ m < 10) and n (1 ≤ n ≤ 10). The next line will contain m integers (from 1 to 9) separated by spaces. These integers form the set S as described above. These integers will be distinct and given in ascending order.
Output
For each case, print the case number and the number of valid n-digit integers in a single line.
Sample Input
3
3 2
1 3 6
3 2
1 2 3
3 3
1 4 6
Sample Output
Case 1: 5
Case 2: 9
Case 3: 9
Note
For the first case the valid integers are
11
13
31
33
66
题意分析：
给出m个数，要求组合成n位数，且相邻两位之间的差不超过2.
解题思路：
直接深搜，对于每一位只考虑前一位的相差2以内的数。
#include <stdio.h>
#include <math.h>
#include <algorithm>
#define N 15
using namespace std;
int a[N];
int n, m, ans;
void dfs(int x, int len)
{
if(len==n)
{
ans++;
return ;
}
for(int i=x-2; i<=x+2; i++)
{
if(i<0 || i>=m)
continue;
if(abs(a[x]-a[i])<=2)
dfs(i, len+1);
}
return ;
}
int main()
{
int t=0, T;
scanf("%d", &T);
while(T--)
{
scanf("%d%d", &m, &n);
for(int i=0; i<m; i++)
scanf("%d", &a[i]);
ans=0;
for(int i=0; i<m; i++)
dfs(i, 1);
printf("Case %d: %d
", ++t, ans);
}
return 0;
}