过前3个点超时

aoding_buxiangxuele @ 2024-08-28 16:08:10

#include<iostream>
using namespace std;
int n,cnt,ay[15],ans,c2[15];// c2列被占用 
int a[100],b[100];//a表示对角线(x-y)被占用(左上到右下) b表示对角线(x+y)被占用 
void dfs(int x){
    if(cnt==n){
        ans++;
        if(ans<=3){
            for(int i=1;i<=n;i++)if(ay[i])cout<<ay[i]<<' ';
            cout<<endl;
        }
        return;
    }
    for(int i=x;i<=n;i++)
        for(int j=1;j<=n;j++){
            if(!ay[i]&&!c2[j]&&!a[n+i-j]&&!b[i+j]){
                c2[j]=a[n+i-j]=b[i+j]=1;
                cnt++;
                ay[i]=j;//第i行的棋子在第j列 
                dfs(x+1); 
                c2[j]=a[n+i-j]=b[i+j]=0;
                cnt--;
                ay[i]=0;
            }
        } 
}
int main(){
    cin>>n;
    dfs(1); 
    cout<<ans;
    return 0;
}

by _Vistion_ @ 2024-08-29 12:17:49

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int n;
int m1[30], m2[30], m3[30], ans[15], mark=0;
void setvalue(int x, int y, int flag){
    ans[x] = y;
    m1[y] = flag;
    m2[x+y] = flag;
    m3[x-y+n] = flag;
}
void dfs(int step){
    if(step>n){
        mark++;
        if(mark<=3){
            for(int i=1; i<=n; i++){
                cout<<ans[i]<<" ";
            }
            cout<<endl;
        }
        return; 
    }
    for(int j=1; j<=n; j++){
        if(m1[j] || m2[step+j] || m3[step-j+n]) continue;
        setvalue(step,j,1);
        dfs(step+1);
        setvalue(step,j,0);
    }
} 
int main(){
    cin>>n;
    dfs(1);
    cout<<mark;
    return 0;
}

by aoding_buxiangxuele @ 2024-08-29 15:49:14

@YZ_zhang 找到了，是我太笨了，多写了一层行的循环，找了一天，谢谢dalao！orz

by hehe_666 @ 2024-09-06 10:01:38

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;

int a[55], n, k, cnt; 
bool hang[55], lie[55], d1[55], d2[55];

void dfs(int step)
{
    if(step > n)
    {
        if(cnt <= 2)
        {
            for(int i = 1; i <= n; i++)
            {
                cout << a[i] << " ";
            }
            cout << endl;
        }
        cnt++;

    }
    for(int i = 1; i <= n; i++)
    {
        int x = step, y = i;
        if(hang[x] || lie[y] || d1[x - y + n] || d2[x + y]) continue;
        hang[x] = lie[y] = d1[x - y + n] = d2[x + y] = true;
        a[step] = i;
        dfs(step + 1);
        hang[x] = lie[y] = d1[x - y + n] = d2[x + y] = false;
    }
}

int main()
{
    cin >> n;
    k = 3;
    dfs(1);
    cout << cnt;
    return 0;
}