解題區/解題區 - UVa

UVa 932 - Checking the N-Queens Problem

contents

  1. 1. Problem
  2. 2. Sample Input
  3. 3. Sample Output
  4. 4. Solution

Problem

給定一個 N * N 的棋盤，檢查 N 個皇后是否衝突，如果存有衝突，嘗試移動一個皇后 (根據他所在的位置進行移動)，如果可以有解則輸出，反之輸出找不到。

Sample Input

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5
00X00
X0000
000X0
0X000
0000X
5
0X000
X0000
000X0
0X000
0000X

Sample Output

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YES
NO
YES
00X00
X0000
000X0
0X000
0000X

Solution

窮舉所有移動可能，並且進行檢查即可。

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#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
int check(int qx[], int qy[], int n) {
    static int used[32] = {}, testcase = 0;
    testcase++;
    for (int i = 0; i < n; i++) {
        if (used[qx[i]] == testcase) return 0;
        used[qx[i]] = testcase;
    }
    testcase++;
    for (int i = 0; i < n; i++) {
        if (used[qy[i]] == testcase) return 0;
        used[qy[i]] = testcase;
    }
    for (int i = 0; i < n; i++) {
        for (int j = i+1; j < n; j++) {
            if (abs(qx[i] - qx[j]) == abs(qy[i] - qy[j]))
                return 0;
        }
    }
    return 1;
}
int main() {
    int n, cases = 0, tx, ty;
    char g[32][32];
    const int dx[] = {0, 0, 1, 1, 1, -1, -1, -1};
    const int dy[] = {1, -1, 0, 1, -1, 0, 1, -1};
    while (scanf("%d", &n) == 1) {
        if (cases++)	puts("");
        for (int i = 0; i < n; i++)
            scanf("%s", g[i]);
        int qx[32], qy[32], m = 0;
        for (int i = 0; i < n;i++) {
            for (int j = 0; j < n; j++) {
                if (g[i][j] == 'X')
                    qx[m] = i, qy[m] = j, m++;
            }
        }
        if (check(qx, qy, m))
            puts("YES");
        else {
            puts("NO");
            int ok = 0;
            for (int i = 0; i < m && !ok; i++) {
                for (int j = 0; j < 8 && !ok; j++) {
                    tx = qx[i] + dx[j], ty = qy[i] + dy[j];
                    while (true) {
                        if (tx < 0 || ty < 0 || tx >= n || ty >= n)
                            break;
                        if (g[tx][ty] == '0') {
                            swap(tx, qx[i]), swap(ty, qy[i]);
                            if (check(qx, qy, m)) {
                                puts("YES");
                                ok = 1;
                                swap(g[qx[i]][qy[i]], g[tx][ty]);
                                for (int i = 0; i < n; i++)
                                    puts(g[i]);
                                break;
                            }
                            swap(tx, qx[i]), swap(ty, qy[i]);
                        }
                        tx += dx[j], ty += dy[j];
                    }
                }
            }
            if (!ok)
                puts("NO");
        }
    }
    return 0;
}
/*
5
00X00
X0000
000X0
0X000
0000X
5
0X000
X0000
000X0
0X000
0000X
*/