Acwing 正方形数组的数目(dfs去重)

解题代码

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
#define rep(i,a,n) for(int i = a; i <= n; i++)
const int N = 20;

int a[N];
int n,ans;
int st[N];

int check(int x){
    int y = sqrt(x);
    return y * y == x;
}

void dfs(int u, int last){
    if(u == n){
        ans++;
    }else{
        rep(i,0,n-1){
            if(st[i])continue;//如果这个数已经被搜过了，就代表不用搜了
            if(i && !st[i - 1] && a[i] == a[i - 1])continue;
            //首先比第一个数大,
            //然后处于递归的同一层，前一个肯定已经恢复现场了，
            //最后判断是否和前一个相等.
            if(!check(a[i] + last)) continue;//不满足条件

            st[i] = 1;
            dfs(u + 1,a[i]);
            st[i] = 0;
        }
    }
}

int main(){
    cin >> n;
    rep(i,0,n - 1){
        cin >> a[i];
    }
    sort(a,a + n);
    rep(i,0,n -1){
        if(!i|| a[i] != a[i - 1]){
            st[i] = 1;

            dfs(1,a[i]);//因为我们这里已经自己选择了一个数作为第一个元素，所以这里的1代表第二层
            //所以递归的出口就是 u == n
            st[i] = 0;
        }
    }
    cout << ans << endl;
    return 0;
}

---

dfs去重输出

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
#define rep(i,a,n) for(int i = a; i <= n; i++)
const int N = 20;

int a[N];
int n,ans;
int st[N];
int path[N];
void dfs(int u, int last){
    if(u == n){
        rep(i,0,n-1) cout<<path[i]<<" ";
        puts("");
    }else{
        rep(i,0,n-1){
            if(st[i])continue;
            if(i && !st[i - 1] && a[i] == a[i - 1])continue;
            st[i] = 1;
            path[u] = a[i];
            dfs(u + 1,a[i]);
            st[i] = 0;
        }
    }
}
int main(){
    cin >> n;
    rep(i,0,n - 1) cin >> a[i];
    sort(a,a + n);
    rep(i,0,n -1){
        if(!i|| a[i] != a[i - 1]){
            st[i] = 1;
            path[0] = a[i];
            dfs(1,a[i]);
            st[i] = 0;
        }
    }
    return 0;
}

结果参考