数据结构 八皇后问题 (暴力，回溯)

Description

设计程序完成如下要求：在8×8的国际象棋棋盘上，放置8个皇后，使得这8个棋子不能互相被对方吃掉。要求：依次输出各种成功的放置方法。（按照字典序输出）

Input

输入包含多组数据，第一行为数据组数，以后每行一个整数k，代表需要输出的字典序第k大的摆放方法。

Output

每组数据一行，依次输出第i列的皇后所在行数。

Sample Input

1

1

Sample Output

15863724

HINT

Append Code

析：这个题同样是暴力回溯，我们可以提前把表都打出来，set来排一下序，最后输出就好，在回溯时，要同时记录每一行，每一列和两个对角线，

可以用一个两维数组vis来记录。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;
 
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5e4 + 5;
const LL mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
bool vis[3][200];
char s[10];
vector<string> ans;
 
void dfs(int cur){
    if(cur == n){ ans.push_back(s); return ;  }
    for(int i = 0; i < n; ++i){
        if(!vis[0][i] && !vis[1][cur+i] && !vis[2][cur-i+n]){
            s[cur] = i + '1';
            vis[0][i] = vis[1][cur+i] = vis[2][cur-i+n] = 1;
            dfs(cur+1);
            vis[0][i] = vis[1][cur+i] = vis[2][cur-i+n] = 0;
        }
    }
}
 
int main(){
    n = 8;
    memset(s, 0, sizeof s);
    memset(vis, 0, sizeof vis);
    dfs(0);
    sort(ans.begin(), ans.end());
    int T;   cin >> T;
    while(T--){
        scanf("%d", &n);
        cout << ans[n-1] << endl;
    }
    return 0;
}