f[i][S] 表示当前集合为 S，最后一个数为 i 的最优解
f[i][S] += f[j][S - i] (j, i ∈ S && j != i && abs(a[i] - a[j]) > k)
——代码
1 #include <cstdio> 2 #include <iostream> 3 #define LL long long 4 5 int a[17]; 6 int n, m, k; 7 LL ans, f[17][1 << 17]; 8 9 inline int read() 10 { 11 int x = 0, f = 1; 12 char ch = getchar(); 13 for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1; 14 for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0'; 15 return x * f; 16 } 17 18 inline int abs(int x) 19 { 20 return x < 0 ? -x : x; 21 } 22 23 int main() 24 { 25 int i, j, S; 26 n = read(); 27 k = read(); 28 m = (1 << n) - 1; 29 for(i = 1; i <= n; i++) a[i] = read(); 30 for(i = 1; i <= n; i++) f[i][1 << i - 1] = 1; 31 for(S = 1; S <= m; S++) 32 for(i = 1; i <= n; i++) 33 if((1 << i - 1) & S) 34 for(j = 1; j <= n; j++) 35 if(i ^ j && (1 << j - 1) & S && abs(a[i] - a[j]) > k) 36 f[i][S] += f[j][(1 << i - 1) ^ S]; 37 for(i = 1; i <= n; i++) ans += f[i][m]; 38 printf("%lld ", ans); 39 return 0; 40 }