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.
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.
1103000509002109400000704000300502006060000050700803004000401000009205800804000107Sample Output
143628579572139468986754231391542786468917352725863914237481695619275843854396127題意
給出一個16*16矩陣的部分格，其中0為空格，要求填充這些空格。
使矩陣滿足橫豎和九個3*3的方格內的數字都包含1~9這9個數字。
思路
很基礎的DFS，先存儲當前每行每列和每塊的數字狀態，然后從左上角開始搜索，遇到零的時候向該點填充一個當前行、當前列、當前塊都不存在的一個數字，若無法填充，則結束此層DFS，若可行，繼續搜索下一層，直到搜索到右下角，標記已經找到答案，此時結束所有DFS，注意在結束的過程中要保留當前所填充的矩陣。
AC 代碼
#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<iostream>using namespace std;int mapp[9][9]; //存儲當前狀態bool isx[9][10],isy[9][10],iss[3][3][10],flag; //行、列、塊判斷void dfs(int y){ if(y>=9) //最后一個數是0的情況下若遞歸到這里，則說明全部填充完畢 { flag=true; return; } for(int i=y; i<9; i++) //從第y行開始，遍歷剩下的點 for(int j=0; j<9; j++) { if(mapp[i][j]==0) //一個需要判斷的點 { bool isc=false; //這一點是否可以被填充 for(int k=1; k<=9; k++) //1-9 if(!isx[j][k]&&!isy[i][k]&&!iss[i/3][j/3][k]) //三種情況都滿足 { isc=true; //假設可以填充 mapp[i][j]=k; //填充的數是k isx[j][k]=isy[i][k]=iss[i/3][j/3][k]=true; //標記該位置 dfs(j!=8?y:y+1); if(flag)return; //找到答案返回，放在這里可以防止在層層結束遞歸的時候狀態被還原 else isc=false; //下層遞歸失敗，標記該點不能填充 mapp[i][j]=0; //還原狀態 isx[j][k]=isy[i][k]=iss[i/3][j/3][k]=false; } if(!isc)return; } else if(i==8&&j==8) //最后一個點不是0的情況 { flag=true; return; } }}int main(){ int n; scanf("%d",&n); for(int ni=0; ni<n; ni++) { memset(isx,false,sizeof(isx)); memset(isy,false,sizeof(isy)); memset(iss,false,sizeof(iss)); flag=false; for(int i=0; i<9; i++) for(int j=0; j<9; j++) { scanf("%1d",&mapp[i][j]); if(mapp[i][j]) { isx[j][mapp[i][j]]=true; isy[i][mapp[i][j]]=true; iss[i/3][j/3][mapp[i][j]]=true; } } dfs(0); for(int i=0; i<9; i++) for(int j=0; j<9; j++) printf(j!=8?"%d":"%d/n",mapp[i][j]); } return 0;}
