P16473 [GKS 2013 #B] Sudoku Checker题解

本题纯烦。

鉴于数据范围比较小，按题意模拟即可。

依次检查每一行，每一列和每一个子矩阵是否符合要求。

注意多测和题目要求的格式化输出。

AC Code

cpp 复制代码

#include <bits/stdc++.h>
#define int long long
#define endl "\n"
using namespace std;

const int MAXN = 45;
int tmp, n, a[MAXN][MAXN];
bool vis[MAXN];

bool check() {
    int sz = n;
    int blk = tmp;
    // 1. 检查每一行
    for (int i = 1; i <= sz; i++) {
        memset(vis, 0, sizeof(vis));
        for (int j = 1; j <= sz; j++) {
            int num = a[i][j];
            if (vis[num]) return false;
            vis[num] = true;
        }
    }
    // 2. 检查每一列
    for (int j = 1; j <= sz; j++) {
        memset(vis, 0, sizeof(vis));
        for (int i = 1; i <= sz; i++) {
            int num = a[i][j];
            if (vis[num]) return false;
            vis[num] = true;
        }
    }
    // 3. 检查每一个子矩阵
    for (int i = 1; i <= sz; i += blk) {
        for (int j = 1; j <= sz; j += blk) {
            memset(vis, 0, sizeof(vis));
            for (int dx = 0; dx < blk; ++dx) {
                for (int dy = 0; dy < blk; ++dy) {
                    int x = i + dx;
                    int y = j + dy;
                    int num = a[x][y];
                    if (vis[num]) return false;
                    vis[num] = true;
                }
            }
        }
    }
    return true;
}

void solve(int case_id) {
    cout << "Case #" << case_id << ": ";
    memset(a, 0, sizeof(a));
    cin >> tmp;
    n = tmp * tmp;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            cin >> a[i][j];
            if (a[i][j] < 1 || a[i][j] > n) {
                for (int k = j + 1; k <= n; k++) cin >> a[i][k];
                for (int k = i + 1; k <= n; k++)
                    for (int l = 1; l <= n; l++) cin >> a[k][l];
                cout << "No\n";
                return;
            }
        }
    }
    if (check()) cout << "Yes\n";
    else cout << "No\n";
}

signed main() {
    cin.tie(0) -> sync_with_stdio(0);
    int t; cin >> t;
    for (int i = 1; i <= t; i++) solve(i);
    return 0;
}

记得三连哦～