ACM学习历程—UESTC 1222 Sudoku（矩阵）(2015CCPC H)

题目链接：http://acm.uestc.edu.cn/#/problem/show/1226

题目大意就是构造一个行列和每个角的2*2都是1234的4*4矩阵。

用dfs暴力搜索，不过需要每一步进行判断是否已经出现了重复，如果最后再判断的话复杂度有点高。

代码:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <string>
#define LL long long

using namespace std;

char str[5][5];
int a[5][5];
bool vis[5];
bool flag;

bool judge(int x[][5])
{
    for (int i = 0; i < 4; ++i)
    {
        memset(vis, false, sizeof(vis));
        for (int j = 0; j < 4; ++j)
        {
            if (!x[i][j]) continue;
            if (vis[x[i][j]]) return false;
            else vis[x[i][j]] = true;
        }
    }
    for (int j = 0; j < 4; ++j)
    {
        memset(vis, false, sizeof(vis));
        for (int i = 0; i < 4; ++i)
        {
            if (!x[i][j]) continue;
            if (vis[x[i][j]]) return false;
            else vis[x[i][j]] = true;
        }
    }
    for (int i = 0; i <= 2; i += 2)
    {
        for (int j = 0; j <= 2; j += 2)
        {
            memset(vis, false, sizeof(vis));
            for (int xx = 0; xx <= 1; xx++)
            {
                for (int yy = 0; yy <= 1; yy++)
                {
                    if (!x[i+xx][j+yy]) continue;
                    if (vis[x[i+xx][j+yy]]) return false;
                    else vis[x[i+xx][j+yy]] = true;
                }
            }
        }
    }
    return true;
}

void input()
{
    for (int i = 0; i < 4; ++i)
    {
        scanf("%s", str[i]);
        for (int j = 0; j < 4; ++j)
        {
            if (str[i][j] == ‘*‘)
                a[i][j] = 0;
            else
                a[i][j] = str[i][j]-‘0‘;
        }
    }
}

void dfs(int i, int j)
{
    if (flag) return;
    i += j/4;
    j %= 4;
    if (i == 4)
    {
        if (judge(a))
        {
            for (int i = 0; i < 4; ++i)
            {
                for (int j = 0; j < 4; ++j)
                    printf("%d", a[i][j]);
                printf("\n");
            }
            flag = true;
        }
        return;
    }
    if (!judge(a)) return;
    if (!a[i][j])
    {
        for (int x = 1; x <= 4; ++x)
        {
            a[i][j] = x;
            dfs(i, j+1);
            a[i][j] = 0;
        }
    }
    else dfs(i, j+1);
}

void work()
{
    flag = false;
    dfs(0, 0);
}

int main()
{
    //freopen("test.in", "r", stdin);
    int T;
    scanf("%d", &T);
    for (int times = 1; times <= T; ++times)
    {
        printf("Case #%d:\n", times);
        input();
        work();
    }
    return 0;
}