HDU 5547.Sudoku

312


题目



点击显/隐题目

Yi Sima was one of the best counselors of Cao Cao. He likes to play a funny game himself. It looks like the modern Sudoku, but smaller.

Actually, Yi Sima was playing it different. First of all, he tried to generate a $4×4$ board with every row contains 1 to 4, every column contains 1 to 4. Also he made sure that if we cut the board into four $2×2$ pieces, every piece contains 1 to 4.

Then, he removed several numbers from the board and gave it to another guy to recover it. As other counselors are not as smart as Yi Sima, Yi Sima always made sure that the board only has one way to recover.

Actually, you are seeing this because you've passed through to the Three-Kingdom Age. You can recover the board to make Yi Sima happy and be promoted. Go and do it!!!


The first line of the input gives the number of test cases, $T(1≤T≤100)$. $T$ test cases follow. Each test case starts with an empty line followed by 4 lines. Each line consist of 4 characters. Each character represents the number in the corresponding cell (one of '1', '2', '3', '4'). '*' represents that number was removed by Yi Sima.

It's guaranteed that there will be exactly one way to recover the board.


For each test case, output one line containing Case #x:, where $x$ is the test case number (starting from 1). Then output 4 lines with 4 characters each. indicate the recovered board.


3
****
2341
4123
3214
*243
*312
*421
*134
*41*
**3*
2*41
4*2*


Case #1:
1432
2341
4123
3214
Case #2:
1243
4312
3421
2134
Case #3:
3412
1234
2341
4123




题解


填数独,*为待填项,共有1,2,3,4四种数字,棋盘大小为4*4,需要满足每行每列以及4个2*2的小区域正好是1,2,3,4


直接爆搜会TLE
注意到 It's guaranteed that there will be exactly one way to recover the board.
有且仅有一组解
可以根据每一行每一列每个正方形区域内已有的数字来推测需要填的数字

具体做法就是记录区域内已有的数字,然后给待填项标记上还能填的数字,只到所有数字都只有一种填法



代码


点击显/隐代码
#include <cstdio>
#include <cstring>
using namespace std;

char m[5][5];
bool vis[5][5][5];
int p[4][2];

void output() {
    for (int i = 0; i < 4; ++i) {
        for (int j = 0; j < 4; ++j)
            if (m[i][j] == '*') {
                printf(" %d%d%d%d", vis[i][j][1], vis[i][j][2], vis[i][j][3],
                       vis[i][j][4]);
            } else
                printf("%5d", m[i][j]);
        printf("\n");
    }
    printf("\n");
}

bool _Change() {
    bool change = false;
    bool has[5];
    memset(has, false, sizeof(has));

    // for (int i = 0; i < 4; ++i) {
    //     printf("(%d %d) ", p[i][0], p[i][1]);
    // }
    // printf("\n");

    for (int i = 0; i < 4; ++i) {
        int x = p[i][0], y = p[i][1];
        if (m[x][y] != '*')
            has[(int)m[x][y]] = true;
    }

    for (int j = 1; j <= 4; ++j) {
        if (has[j] == true) {
            for (int i = 0; i < 4; ++i) {
                int x = p[i][0], y = p[i][1];
                if (m[x][y] == '*' && vis[x][y][j] == false) {
                    change = true;
                    vis[x][y][j] = true;
                }
            }
        }
    }

    for (int i = 0; i < 4; ++i) {
        int x = p[i][0], y = p[i][1];
        if (m[x][y] == '*') {
            int ok = 0;
            for (int j = 1; j <= 4; ++j) {
                if (vis[x][y][j] == false) {
                    if (ok == 0) {
                        ok = j;
                    } else {
                        ok = 0;
                        break;
                    }
                }
            }
            if (ok != 0)
                m[x][y] = ok;
        }
    }
    return change;
}

bool Change() {
    bool ok = false;
    for (int i = 0; i < 4; ++i) {
        for (int j = 0; j < 4; ++j) {
            p[j][0] = i;
            p[j][1] = j;
        }
        ok |= _Change();
    }
    for (int i = 0; i < 4; ++i) {
        for (int j = 0; j < 4; ++j) {
            p[j][0] = j;
            p[j][1] = i;
        }
        ok |= _Change();
    }
    for (int i = 0; i < 2; ++i) {
        for (int j = 0; j < 2; ++j) {
            p[i * 2 + j][0] = i;
            p[i * 2 + j][1] = j;
        }
    }
    ok |= _Change();

    for (int i = 0; i < 2; ++i) {
        for (int j = 0; j < 2; ++j) {
            p[i * 2 + j][0] = i + 2;
            p[i * 2 + j][1] = j;
        }
    }
    ok |= _Change();

    for (int i = 0; i < 2; ++i) {
        for (int j = 0; j < 2; ++j) {
            p[i * 2 + j][0] = i;
            p[i * 2 + j][1] = j + 2;
        }
    }
    ok |= _Change();

    for (int i = 0; i < 2; ++i) {
        for (int j = 0; j < 2; ++j) {
            p[i * 2 + j][0] = i + 2;
            p[i * 2 + j][1] = j + 2;
        }
    }
    ok |= _Change();

    return ok;
}

int main() {
    int T;
    scanf("%d", &T);
    for (int kase = 1; kase <= T; ++kase) {
        for (int i = 0; i < 4; ++i)
            scanf("%s", m[i]);

        for (int i = 0; i < 4; ++i)
            for (int j = 0; j < 4; ++j)
                if (m[i][j] == '*')
                    memset(vis[i][j], false, sizeof(vis[i][j]));
                else
                    m[i][j] = m[i][j] - '0';

        while (Change()) ;
        printf("Case #%d:\n", kase);
        for (int i = 0; i < 4; ++i) {
            for (int j = 0; j < 4; ++j)
                printf("%d", m[i][j]);
            printf("\n");
        }
    }
    return 0;
}
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