POJ 2676 深搜+剪枝

Sudoku

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 23467		Accepted: 10979		Special Judge

Description

Sudoku is a very simple task. A square table with 9 rows and 9 columns is divided to 9 smaller squares 3x3 as shown on the Figure. In some of the cells are written decimal digits from 1 to 9. The other cells are empty. The goal is to fill the empty cells with decimal digits from 1 to 9, one digit per cell, in such way that in each row, in each column and in each marked 3x3 subsquare, all the digits from 1 to 9 to appear. Write a program to solve a given Sudoku-task. 

Input

The input data will start with the number of the test cases. For each test case, 9 lines follow, corresponding to the rows of the table. On each line a string of exactly 9 decimal digits is given, corresponding to the cells in this line. If a cell is empty it is represented by 0.

Output

For each test case your program should print the solution in the same format as the input data. The empty cells have to be filled according to the rules. If solutions is not unique, then the program may print any one of them.

Sample Input

1
103000509
002109400
000704000
300502006
060000050
700803004
000401000
009205800
804000107

Sample Output

143628579
572139468
986754231
391542786
468917352
725863914
237481695
619275843
854396127

解题思路：直接深搜会搜索很多错误的情况，在搜索时，将这些情况排除

#include <stdio.h>
#include <iostream>
#include <string>
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)

using namespace std;

int n;
int map_hang[10][10];
int map_lie[10][10];
int map_gezi[4][4][10];
int shudu[10][10];

bool hang_lie_gezi(int x, int y, int v) {
    return map_lie[y][v] == 0 && map_hang[x][v] == 0 && map_gezi[(x+2)/3][(y+2)/3][v] == 0;
}//用于检测v值是否可以填在该位置,可以滤过0 0
bool empty(int x, int y) {
    return !shudu[x][y];
}//用于检测坐标（X，Y）是否有数

void add(int x, int y, int v) {
    shudu[x][y] = v;

    map_hang[x][v]=1; //此处应为=1,++是为了debug
    map_lie[y][v]=1;  //此处应为=1,++是为了debug

    map_gezi[(x + 2) / 3][(y + 2) / 3][v]=1;
}
void reback(int lx, int ly, int lv) {
    shudu[lx][ly] = 0;

    map_hang[lx][lv]=0; //此处应为=0,--是为了debug
    map_lie[ly][lv]=0;  //此处应为=0,--是为了debug

    map_gezi[(lx + 2) / 3][(ly + 2) / 3][lv]=0;
}

bool dfs(int x, int y) {
    if (x==10 || y == 10)
    {
        return 1;
    }

    int flag = 0;

    if (shudu[x][y])
    {
        if (y == 9)
            flag = dfs(x + 1, 1);
        else
            flag = dfs(x, y + 1);

        return flag;
    }
    else
    {
        for (int v = 1; v <= 9; v++) {
            if (hang_lie_gezi(x, y, v))
            {
                add(x, y, v);

                if (y == 9)
                    flag = dfs(x + 1, 1);
                else
                    flag = dfs(x, y + 1);

                if (flag == 0) {
                    reback(x, y, v);
                }
                else
                {
                    return 1;
                }
            }
        }
    }
    return 0;
}

string s;

int main() {
    fio;
    cin >> n;
    while (n--)
    {
        memset(shudu, 0, sizeof shudu);
        memset(map_hang, 0, sizeof map_hang);
        memset(map_gezi, 0, sizeof map_gezi);
        memset(map_lie, 0, sizeof map_lie);

        for (int i = 1; i < 10; i++) {
            cin >> s;
            for (int j = 1; j < 10; j++) {
                shudu[i][j] = s[j-1]-'0';
            }
        }
        for (int i = 1; i < 10; i++) {
            for (int j = 1; j < 10; j++) {
                if (shudu[i][j])
                {
                    add(i, j, shudu[i][j]);
                }
            }
        }

        dfs(1, 1);

        for (int i = 1; i <= 9; i++) {
            for (int j = 1; j <= 9; j++) {
                cout << shudu[i][j];
            }
            cout << endl;
        }
    }
    return 0;
}