ACM : POJ 2676 SudoKu DFS - 数独

最新推荐文章于 2022-07-18 20:29:33 发布

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最新推荐文章于 2022-07-18 20:29:33 发布

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原文链接：http://www.cnblogs.com/HDMaxfun/p/5869199.html

本文介绍了一种解决数独问题的算法实现，采用深度优先搜索（DFS）策略填充空缺单元格，确保每一行、每一列及每个3x3宫格内的数字1到9各出现一次。

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SudoKu

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

POJ 2676

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

Sample Output

/*/
数独

AC代码：
/*/

#include"iostream"
#include"cstdio"
#include"cstring"
#include"queue"
#include"string"
using namespace std;
int map[11][11],count,n;
bool F[11][11];
bool H[11][11];
bool L[11][11];

struct Node {
	int x,y;
} node[100];

int MAP(int i,int j) {
	if(i<=3&&j<=3)return 1;
	else if(i<=3&&j<=6)return 2;
	else if(i<=3&&j<=9)return 3;
	else if(i<=6&&j<=3)return 4;
	else if(i<=6&&j<=6)return 5;
	else if(i<=6&&j<=9)return 6;
	else if(i<=9&&j<=3)return 7;
	else if(i<=9&&j<=6)return 8;
	else if(i<=9&&j<=9)return 9;
}

int  DFS(int n) {
	if(n>count) {
		return 1;
	}
	for(int j=1; j<=9; j++) {
		if(!H[node[n].x][j]&&!L[node[n].y][j]&&!F[MAP(node[n].x,node[n].y)][j]) {
			H[node[n].x][j]=L[node[n].y][j]=F[MAP(node[n].x,node[n].y)][j]=1;
			map[node[n].x][node[n].y]=j;
			if(DFS(n+1))return 1;
			H[node[n].x][j]=L[node[n].y][j]=F[MAP(node[n].x,node[n].y)][j]=0;
		}
	}
	return 0;
}


int main() {
	int flag=0;
	while(cin>>n) {
		for(int i=1; i<=n; i++) {
			memset(map,0,sizeof(map));
			memset(H,0,sizeof(H));
			memset(F,0,sizeof(F));
			memset(L,0,sizeof(L));
			count = 0;
			for(int j=1; j<=9; j++)
				for(int k=1; k<=9; k++) {
					scanf("%1d",&map[j][k]);
					if(map[j][k]==0) {
						count++;
						node[count].x=j;
						node[count].y=k;
					} else {
						H[j][map[j][k]]=1;
						L[k][map[j][k]]=1;
						F[MAP(j,k)][map[j][k]]=1;
					}
				}
			DFS(1);
			for(int j=1; j<=9; j++) {
				for(int k=1; k<=9; k++)
					cout<<map[j][k];
				cout<<endl;
			}

		}
	}
	return 0;
}

转载于:https://www.cnblogs.com/HDMaxfun/p/5869199.html