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

传送门

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

 #include <cstdio>
 #include <iostream>
 #define LL long long

 int a[];
 int n, m, k;
 LL ans, f[][ << ];

 inline int read()
 {
     int x = , f = ;
     char ch = getchar();
     for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -;
     for(; isdigit(ch); ch = getchar()) x = (x << ) + (x << ) + ch - '';
     return x * f;
 }

 inline int abs(int x)
 {
     return x <  ? -x : x;
 }

 int main()
 {
     int i, j, S;
     n = read();
     k = read();
     m = ( << n) - ;
     for(i = ; i <= n; i++) a[i] = read();
     for(i = ; i <= n; i++) f[i][ << i - ] = ;
     for(S = ; S <= m; S++)
         for(i = ; i <= n; i++)
             if(( << i - ) & S)
                 for(j = ; j <= n; j++)
                     if(i ^ j && ( << j - ) & S && abs(a[i] - a[j]) > k)
                         f[i][S] += f[j][( << i - ) ^ S];
     for(i = ; i <= n; i++) ans += f[i][m];
     printf("%lld\n", ans);
     return ;
 }