1128 N Queens Puzzle (20分)

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 (Q​1​​,Q​2​​,⋯,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 Q​1​​ Q​2​​ ... 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:

4
8 4 6 8 2 7 1 3 5
9 4 6 7 2 8 1 9 5 3
6 1 5 2 6 4 3
5 1 3 5 2 4

Sample Output:

YES
NO
NO
YES

思路： 

把每一行每一列，每一右下线，左下线分别标记出来即可。

AC代码：

#include <bits/stdc++.h>
#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <time.h>
#include <string.h>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define sdddd(x,y,z,k) scanf("%d%d%d%d", &x, &y, &z, &k)
#define sddd(x,y,z) scanf("%d%d%d", &x, &y, &z)
#define sdd(x,y) scanf("%d%d", &x, &y)
#define sd(x) scanf("%d", &x)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
//#define mp Debug(x) printf("%d\n", &x);
#define pb push_back
#define ms(x, y) memset(x, y, sizeof x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll MOD = 1046513837;
const int maxn = 1e5 + 50;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
//typedef vector<ll> vec;
//typedef vector<vec> mat;
template <class T>
inline bool scan_d(T &ret) {
	char c; int sgn;
	if (c = getchar(), c == EOF) return 0;
	while (c != '-' && (c<'0' || c>'9')) c = getchar();
	sgn = (c == '-') ? -1 : 1;
	ret = (c == '-') ? 0 : (c - '0');
	while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
	ret *= sgn;
	return 1;
}

int n, m;
int col[1005], row[1005];
int rightDown[1005][1005];
int leftDown[1005][1005];
vector<pair<int, int> > vec;
int main() {
    int tmp;
    cin >> n;
    rep(i, 1, n){
        cin >> m;
        int flag = 1;
        rep(j, 1, 1000){
            col[j] = 0;
            row[j] = 0;
            rightDown[1][j] = 0;
            rightDown[j][m] = 0;
            leftDown[j][m] = 0;
            leftDown[m][j] = 0;
        }
        rep(j, 1, m){
            cin >> tmp;
            row[j]++;
            if(row[j] > 1) flag = 0;
            col[tmp]++;
            if(col[tmp] > 1) flag = 0;
            int k = min(j-1, m-tmp);
            rightDown[j-k][tmp+k]++;
            if(rightDown[j-k][tmp+k] > 1)
                flag = 0;
            k = min(m-j, m-tmp);
            leftDown[j+k][tmp+k]++;
            if(leftDown[j+k][tmp+k] > 1)
                flag = 0;
        }
        printf("%s\n",flag?"YES":"NO");
    }
	return 0;
}