hud 5547 sodoku (DFS)

题目连接

题目大意：4*4的网格做数独，*代表空的格子，要求将4*4分解为4个2*2的网格，每个2*2网格也满足数独（包含1234）

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int cnt[4];
bool vis[4][5];
char m[4][4];
bool dfs(int x,int y){
    int nx,ny;
    nx = x/2;
    ny = y/2;
    if(x > 3){
        return true;
    }
    if(m[x][y] != '*'){
        if((x == 1 && y == 1) || (x == 1 && y == 3) ||(x == 3 && y == 1)||(x == 3 && y == 3)){
                if(m[0+2*nx][0+2*ny] + m[1+2*nx][1+2*ny] + m[1+2*nx][0+2*ny] + m[0+2*nx][1+2*ny] != 10 + 4 * '0' ){
                    return false;
                }
                else{
                    if(m[0+2*nx][0+2*ny] == m[1+2*nx][1+2*ny] || m[1+2*nx][0+2*ny] == m[0+2*nx][1+2*ny]) return false;
                }
        }
        if(y + 1 < 4) return dfs(x,y+1);
        else  return dfs(x+1,0);
    }
    else if(m[x][y] == '*'){
        for(int i = 1;i <= 4;i ++){
            int flag = 1;
            m[x][y] = i + '0';
            for(int j = 0;j < 4;j ++){
                if(j == y) continue;
                if(m[x][j] - '0' == i) {flag = 0;break;}
            }
            for(int j = 0;j < 4;j ++){
                if(x == j) continue;
                if(m[j][y] - '0' == i) {flag = 0;break;}
            }
            if((x == 1 && y == 1) || (x == 1 && y == 3) ||(x == 3 && y == 1)||(x == 3 && y == 3)){
                if(m[0+2*nx][0+2*ny] + m[1+2*nx][1+2*ny] + m[1+2*nx][0+2*ny] + m[0+2*nx][1+2*ny] != 10 + 4 * '0'){
                    flag = 0;
                }
                else{
                    if(m[0+2*nx][0+2*ny] == m[1+2*nx][1+2*ny] || m[1+2*nx][0+2*ny] == m[0+2*nx][1+2*ny])  flag = 0;
                }
            }
            if(flag){
                m[x][y] = i + '0';
                if(y + 1 < 4) {
                    if(dfs(x,y+1)) return true;
                }
                else {
                    if(dfs(x+1,0)) return true;
                }
            }
            m[x][y] = '*';
        }
        return false;
    }
}
int main()
{
    int cas;
    cin >> cas;
    getchar();
    int cnt = 1;
    while(cas --){
        for(int i = 0;i < 4;i ++){
            for(int j = 0;j < 4;j ++){
                cin >> m[i][j];
            }
            getchar();
        }
        dfs(0,0);
        printf("Case #%d:
",cnt++);
        for(int i = 0;i < 4;i ++){
            for(int j = 0;j < 4;j ++){
                cout << m[i][j];
            }
            cout << endl;
        }
    }
    return 0;
}