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11005: 【原1005】数独

题目

题目描述

author: 吴杰 原OJ链接:https://acm.sjtu.edu.cn/OnlineJudge-old/problem/1005

题目描述

数独游戏,具体规则如下:

每一行都用到1,2,3,4,5,6,7,8,9, 位置不限,

每一列都用到1,2,3,4,5,6,7,8,9, 位置不限,

每3*3的格子(共9个这样的格子)都用到1,2,3,4,5,6,7,8,9, 位置不限,

游戏过程就是用1,2,3,4,5,6,7,8,9填充空白,并满足每行,每列,每个九宫格都用到1,2,3,4,5,6,7,8,9,

如下是个正确的sudoku:

数独

输入格式

输入n个数独,你来验证它是否违反规则。

第一行为数独个数,第二行开始为第一个数独,之后第二个,至第n个。

注意:每个数独间有个回车隔开。

输出格式

若正确则输出"Right",否则输出"Wrong", 输出一个换一行。

说明

1<=n<=20(输入的数独个数)

不论输入的数独是否正确,数据保证每个数都在1-9间。

Sample Input

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

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

Sample Output

Right
Wrong

BugenZhao's solution

//
// Created by BugenZhao on 2019/3/23.
//

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

using ll=long long;

int main() {
    int tmp;
    bool flag[3][9][9];
    int n;
    cin >> n;
    while (n--) {
        bool wrong = false;
        for (int i = 0; i < 9; ++i) {
            for (int j = 0; j < 9; ++j) {
                for (int k = 0; k < 3; ++k) {
                    flag[k][i][j] = false;
                }
            }
        }
        for (int i = 0; i < 9; ++i) {
            for (int j = 0; j < 9; ++j) {
                scanf("%d", &tmp);
                if (wrong) continue;
                int k = i / 3 * 3 + j / 3;
                if (flag[0][i][tmp] || flag[1][j][tmp] || flag[2][k][tmp]) {
                    wrong = true;
                } else {
                    flag[0][i][tmp] = flag[1][j][tmp] = flag[2][k][tmp] = true;
                }
            }
        }
        if (wrong) printf("Wrong\n");
        else printf("Right\n");
    }
    return 0;
}

CallMeInk's solution

#include <iostream>

using namespace std;

int a[10][10];

int main()
{
    int t,a[10][10],i,j;
    cin >> t;
    for(int tt=1;tt<=t;tt++)
    {
        bool flag = true;
        for(i=1;i<=9;i++)
            for(j=1;j<=9;j++)
                cin >> a[i][j];
        for(int i=1;i<=9;i++)
        {
            int sum1 = 0,sum2 = 1;
            for(int j=1;j<=9;j++)
            {
                sum1 += a[i][j];
                sum2 *= a[i][j];
            }
            if (sum1 != 45 || sum2 != 362880)
            {
                flag = false;
                break;
            }
        }
        for(int j=1;j<=9;j++)
        {
            int sum1 = 0,sum2 = 1;
            for(int i=1;i<=9;i++)
            {
                sum1 += a[i][j];
                sum2 *= a[i][j];
            }
            if (sum1 != 45 || sum2 != 362880)
            {
                flag = false;
                break;
            }
        }
        for(int i=1;i<=7;i+=3)
            for(int j=1;j<=7;j+=3)
                {
                    int sum1 = 0,sum2 = 1;
                    for (int x=i;x<=i+2;x++)
                        for(int y=j;y<=j+2;y++)
                            {
                                sum1 += a[x][y];
                                sum2 *= a[x][y];
                            }
                    if (sum1 != 45 || sum2 != 362880)
                    {
                        flag = false;
                        break;
                    }
                }
        if (flag) cout << "Right" << endl;
        else cout << "Wrong" << endl;
    }
    return 0;
}

FineArtz's solution

/* 数独 */
#include <iostream>
using namespace std;

inline short WhichBlock(int x, int y){
    if (x <= 3){
        if (y <= 3) return 1;
        if (y <= 6) return 2;
        if (y <= 9) return 3;
    }
    if (x <= 6){
        if (y <= 3) return 4;
        if (y <= 6) return 5;
        if (y <= 9) return 6;
    }
    if (x <= 9){
        if (y <= 3) return 7;
        if (y <= 6) return 8;
        if (y <= 9) return 9;
    }
}
int main(){
    int k;
    cin >> k;
    for (int data = 0; data != k; ++data){
        bool line[10][10] = {0}, row[10][10] = {0}, block[10][10] = {0}, flag = 1;
        short num = 0;
        for (int i = 0; i != 9; ++i){
            for (int j = 0; j != 9; ++j){
                cin >> num;
                if (line[i][num] || row[j][num] || block[WhichBlock(i + 1, j + 1)][num]){
                    flag = 0;
                    //break;
                }
                line[i][num] = row[j][num] = block[WhichBlock(i + 1, j + 1)][num] = 1;
            }
            //if (!flag) break;
        }
        if (flag) cout << "Right" << endl;
        else cout << "Wrong" << endl;
    }
    return 0;
}

ligongzzz's solution

#include "iostream"
using namespace std;

int main() {
    int num;
    cin >> num;

    for (; num > 0; num--) {
        int data[9][9];
        for (int i = 0; i < 9; i++)
            for (int j = 0; j < 9; j++)
                cin >> data[i][j];

        //Check
        bool flag = false;
        for (int i = 0; i < 9; i++) {
            bool check[9] = { 0 };
            for (int j = 0; j < 9; j++) {
                if (check[data[i][j] - 1]) {
                    cout << "Wrong" << endl;
                    flag = true;
                    break;
                }
                check[data[i][j] - 1] = true;
            }
            if (flag)
                break;
        }
        if (flag)
            continue;

        //Check
        flag = false;
        for (int i = 0; i < 9; i++) {
            bool check[9] = { 0 };
            for (int j = 0; j < 9; j++) {
                if (check[data[j][i] - 1]) {
                    cout << "Wrong" << endl;
                    flag = true;
                    break;
                }
                check[data[j][i] - 1] = true;
            }
            if (flag)
                break;
        }
        if (flag)
            continue;

        flag = false;
        for (int i = 0; i < 9; i += 3) {
            for (int j = 0; j < 9; j += 3) {
                //Small Square
                bool check[9] = { 0 };
                for (int m = 0; m < 3; m++) {
                    for (int n = 0; n < 3; n++) {
                        if (check[data[i + m][j + n] - 1]) {
                            flag = true;
                            cout << "Wrong" << endl;
                            goto outside;
                        }
                        check[data[i + m][j + n] - 1] = true;
                    }
                }
            }
        }

    outside:
        if (flag)
            continue;
        cout << "Right" << endl;
    }

    return 0;
}

Neight99's solution

#include <iostream>

using namespace std;

class sudoku {
   private:
    int **storage;

   public:
    sudoku();
    ~sudoku();
    bool checkline();
    bool checkcolumn();
    bool checknine();
    friend istream &operator>>(istream &, sudoku &);
};

sudoku::sudoku() {
    storage = new int *[9];
    for (int i = 0; i < 9; i++) {
        storage[i] = new int[9];
        for (int j = 0; j < 9; j++) {
            storage[i][j] = 0;
        }
    }
}

sudoku::~sudoku() {
    for (int i = 0; i < 9; i++) {
        delete[] storage[i];
    }
    delete[] storage;
    storage = NULL;
}

bool sudoku::checkline() {
    bool line[9][9] = {0};
    for (int i = 0; i < 9; i++) {
        for (int j = 0; j < 9; j++) {
            int tmp = storage[i][j];
            if (line[i][tmp - 1] == 0) {
                line[i][tmp - 1] = 1;
            } else {
                return 0;
            }
        }
    }
    return 1;
}

bool sudoku::checkcolumn() {
    bool line[9][9] = {0};
    for (int j = 0; j < 9; j++) {
        for (int i = 0; i < 9; i++) {
            int tmp = storage[i][j];
            if (line[tmp - 1][j] == 0) {
                line[tmp - 1][j] = 1;
            } else {
                return 0;
            }
        }
    }
    return 1;
}

bool sudoku::checknine() {
    bool line[9][9] = {0};
    for (int i = 0; i < 9; i++) {
        for (int j = 0; j < 9; j++) {
            int tmp = storage[i][j];
            if (line[(i / 3) * 3 + j / 3][tmp - 1] == 0) {
                line[(i / 3) * 3 + j / 3][tmp - 1] = 1;
            } else {
                return 0;
            }
        }
    }
    return 1;
}

istream &operator>>(istream &is, sudoku &s) {
    for (int i = 0; i < 9; i++) {
        for (int j = 0; j < 9; j++) {
            is >> s.storage[i][j];
        }
    }
    return is;
}

int main() {
    int n;
    sudoku *su;

    cin >> n;
    su = new sudoku[n];
    for (int i = 0; i < n; i++) {
        cin >> su[i];
    }

    for (int i = 0; i < n; i++) {
        bool check =
            (su[i].checkline() && su[i].checkcolumn() && su[i].checknine());
        if (check == 1) {
            cout << "Right" << endl;
        } else {
            cout << "Wrong" << endl;
        }
    }

    return 0;
}

satgo1546's solution

#include <iostream>
using namespace std;

// 要检查的方格组
const int f[3][2][9] = {
    {
        {0, 9, 18, 27, 36, 45, 54, 63, 72},
        {0, 1, 2, 3, 4, 5, 6, 7, 8},
    }, {
        {0, 1, 2, 3, 4, 5, 6, 7, 8},
        {0, 9, 18, 27, 36, 45, 54, 63, 72},
    }, {
        {0, 3, 6, 27, 30, 33, 54, 57, 60},
        {0, 1, 2, 9, 10, 11, 18, 19, 20},
    }
};
int a[81];

int main(int argc, char *argv[]) {
    cin >> argc;
    while (argc--) {
        bool b = true;
        for (int i = 0; i < 81; i++) cin >> a[i];
        for (int i = 0; i - 3; i++) {
            for (int j = 0; j - 9; j++) {
                // 检查方格组中的数字有无重复
                int t = 0;
                for (int k = 0; k - 9; k++) t ^= 1 << a[f[i][0][j] + f[i][1][k]];
                if (t - 1022) b = false;
            }
        }
        cout << (b ? "Right" : "Wrong") << endl;
    }
    return 0;
}

skyzh's solution

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

bool verify(bool data[10]) {
    for (int i = 1; i <= 9; i++) if (!data[i]) return false;
    return true;
}

void run() {
    int D[9][9];
    for (int i = 0; i < 9; i++) {
        for (int j = 0; j < 9; j++) {
            scanf("%d", &D[i][j]);
        }
    }
    bool ok = true;
    for (int i = 0; i < 9; i++) {
        bool data[10] = { 0 };
        for (int j = 0; j < 9; j++) {
            data[D[i][j]] = true;
        }
        ok &= verify(data);
    }
    for (int i = 0; i < 9; i++) {
        bool data[10] = { 0 };
        for (int j = 0; j < 9; j++) {
            data[D[j][i]] = true;
        }
        ok &= verify(data);
    }
    for (int R = 0; R < 3; R++) {
        for (int C = 0; C < 3; C++) {
            bool data[10] = { 0 };
            for (int x = 0; x < 3; x++) {
                for (int y = 0; y < 3; y++) {
                    data[D[R*3+x][C*3+y]] = true;
                }
            }
            ok &= verify(data);
        }
    }
    if (ok) printf("Right\n"); else printf("Wrong\n");
}

int main() {
    int N;
    scanf("%d", &N);
    for (int t = 0; t < N; t++) run();
    return 0;
}

yyong119's solution

#include <iostream>
int n,i,j,k,l;
int a[10][10],c[10];
bool te;
int main(){
    using namespace std;
    cin>>n;
    while(n){
        te=false;
        for (i=1; i<=9; i++){
            for (j=1; j<=9; j++) c[j]=0;
            for (j=1; j<=9; j++){
                    cin>>a[i][j];
                    c[a[i][j]]++;
            }
            for (j=1; j<=9; j++) if (c[j]>1) te=true;
        }
        for (i=1; i<=3; i++)
        for (j=1; j<=3; j++){
            for (k=1; k<=9; k++) c[k]=0;
            for (k=3*i-2; k<=3*i; k++)
            for (l=3*j-2; l<=3*j; l++)
                c[a[k][l]]++;
            for (k=1; k<=9; k++) if (c[k]>1) te=true;
        }
        if (te) cout<<"Wrong\n";else cout<<"Right\n";
        n--;
    }
    return 0;
}