P2915 [USACO08NOV]奶牛混合起来Mixed Up Cows

${\color{Cyan}{>>Question}}$

一道经典的状压

开始没有想出来,但看了培训后,按照$dfs$的思路先想了一遍

假设我们要$dfs$,那我们要知道哪些牛用了,当前是哪头牛

所以状压的状态也就出来了

令$f[i][j]$表示$i$状态,当前在$j$号牛的方案数

有$$f[i][j] = \sum_{k\in (i-\left \{ j \right \})} f[i-\left \{ j \right \}][k]$$

!!初始化!!

有$$f[\left \{ i \right \}][i] = 1$$

上代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define ll long long
using namespace std; 

template <typename T> void in(T &x) {
    x = 0; T f = 1; char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
    while( isdigit(ch)) {x = 10 * x + ch - 48; ch = getchar();}
    x *= f;
}

template <typename T> void out(T x) {
    if(x < 0) x = -x , putchar('-');
    if(x > 9) out(x/10);
    putchar(x%10 + 48);
}
//-------------------------------------------------------

const int N = 18;

int n,m,s[N];
ll f[1<<N][N];

int main() {
    int i,j,k;
    in(n); in(m);
    for(i = 1;i <= n; ++i) in(s[i]),f[1<<(i-1)][i] = 1;//debug 1<<(i-1) -> 0
    for(i = 0;i < (1<<n); ++i) {
        for(j = 1;j <= n; ++j) {
            if(!(i&(1<<(j-1)))) continue;
            for(k = 1;k <= n; ++k) {
                if(!(i&(1<<(k-1))) || k == j || abs(s[j]-s[k]) <= m) continue;
                f[i][j] += f[i^(1<<(j-1))][k];
            }
        }
    }
    ll ans = 0;
    for(i = 1;i <= n; ++i) ans += f[(1<<n)-1][i];
    out(ans);
    return 0;
}