PAT (Advanced Level) Practise 1128 N Queens Puzzle (20)

1128. N Queens Puzzle (20)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

The "eight queens puzzle" is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other. Thus, a solution requires that no two queens share the same row, column, or diagonal. The eight queens puzzle is an example of the more general N queens problem of placing N non-attacking queens on an N×N chessboard. (From Wikipedia - "Eight queens puzzle".)

Here you are NOT asked to solve the puzzles. Instead, you are supposed to judge whether or not a given configuration of the chessboard is a solution. To simplify the representation of a chessboard, let us assume that no two queens will be placed in the same column. Then a configuration can be represented by a simple integer sequence (Q₁, Q₂, ..., Q_N), where Q_i is the row number of the queen in the i-th column. For example, Figure 1 can be represented by (4, 6, 8, 2, 7, 1, 3, 5) and it is indeed a solution to the 8 queens puzzle; while Figure 2 can be represented by (4, 6, 7, 2, 8, 1, 9, 5, 3) and is NOT a 9 queens' solution.

 Figure 1 Figure 2

Input Specification:

Each input file contains several test cases. The first line gives an integer K (1 < K <= 200). Then K lines follow, each gives a configuration in the format "N Q₁ Q₂ ... Q_N", where 4 <= N <= 1000 and it is guaranteed that 1 <= Q_i <= N for all i=1, ..., N. The numbers are separated by spaces.

Output Specification:

For each configuration, if it is a solution to the N queens problem, print "YES" in a line; or "NO" if not.

Sample Input:

48 4 6 8 2 7 1 3 59 4 6 7 2 8 1 9 5 36 1 5 2 6 4 35 1 3 5 2 4

Sample Output:

YESNONOYES

判断一个n皇后是否合法，简单题，枚举行列以及斜线即可。

#include<map>   #include<set>  #include<ctime>    #include<cmath>        #include<queue>     #include<string>    #include<vector>    #include<cstdio>        #include<cstring>      #include<iostream>    #include<algorithm>        #include<functional>    using namespace std;#define ms(x,y) memset(x,y,sizeof(x))        #define rep(i,j,k) for(int i=j;i<=k;i++)        #define per(i,j,k) for(int i=j;i>=k;i--)        #define loop(i,j,k) for (int i=j;i!=-1;i=k[i])        #define inone(x) scanf("%d",&x)        #define intwo(x,y) scanf("%d%d",&x,&y)        #define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z)      #define infou(x,y,z,p) scanf("%d%d%d%d",&x,&y,&z,&p)     #define lson x<<1,l,mid    #define rson x<<1|1,mid+1,r    #define mp(i,j) make_pair(i,j)    #define ft first    #define sd second    typedef long long LL;typedef pair<int, int> pii;const int low(int x) { return x&-x; }const int INF = 0x7FFFFFFF;const int mod = 1e9 + 7;const int N = 1e3 + 50;const double eps = 1e-10;int n, m, x;int f[N][N];int main(){  inone(n);  while (n--)  {    inone(m);    rep(i, 1, m) rep(j, 1, m) f[i][j] = 0;    rep(i, 1, m) inone(x), f[i][x] = 1;    int flag = 1;    rep(i, 1, m)    {      int cnt = 0;      rep(j, 1, m) cnt += f[i][j];      if (cnt != 1) { flag = 0; break; }      rep(j, 1, m) cnt += f[j][i];      if (cnt != 2) { flag = 0; break; }    }    rep(i, 2, m + m)    {      int cnt = 0;      rep(j, max(1, i - m), min(m, i - 1)) cnt += f[j][i - j];      if (cnt > 1) { flag = 0; break; }    }    puts(flag ? "YES" : "NO");  }  return 0;}