傳送門

f[i][S] 表示當前集合為 S，最後一個數為 i 的最優解

f[i][S] += f[j][S - i] (j, i ∈ S && j != i && abs(a[i] - a[j]) > k)

——代碼

 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\n", ans);
39     return 0;
40 }

[luoguP2915] [USACO08NOV]奶牛混合起來Mixed Up Cows（DP）