hdu 5547 Sudoku
2016-03-08 15:59
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Sudoku
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 710 Accepted Submission(s): 268
Problem Description
Yi Sima was one of the best counselors of Cao Cao. He likes to play a funny game himself. It looks like the modern Sudoku, but smaller.
Actually, Yi Sima was playing it different. First of all, he tried to generate a 4×4 board
with every row contains 1 to 4, every column contains 1 to 4. Also he made sure that if we cut the board into four 2×2 pieces,
every piece contains 1 to 4.
Then, he removed several numbers from the board and gave it to another guy to recover it. As other counselors are not as smart as Yi Sima, Yi Sima always made sure that the board only has one way to recover.
Actually, you are seeing this because you've passed through to the Three-Kingdom Age. You can recover the board to make Yi Sima happy and be promoted. Go and do it!!!
Input
The first line of the input gives the number of test cases, T(1≤T≤100). T test
cases follow. Each test case starts with an empty line followed by 4 lines. Each line consist of 4 characters. Each character represents the number in the corresponding cell (one of '1', '2', '3', '4'). '*' represents that number was removed by Yi Sima.
It's guaranteed that there will be exactly one way to recover the board.
Output
For each test case, output one line containing Case #x:, where x is
the test case number (starting from 1). Then output 4 lines with 4 characters each. indicate the recovered board.
Sample Input
3 **** 2341 4123 3214 *243 *312 *421 *134 *41* **3* 2*41 4*2*
Sample Output
Case #1: 1432 2341 4123 3214 Case #2: 1243 4312 3421 2134 Case #3: 3412 1234 2341 4123
Source
The 2015 China Collegiate Programming Contest
Recommend
wange2014
题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=5547
题目大意:4个数字的数独。
解体思路:dfs,枚举填入数字1-4,每次判断填入数字是否符合就行。同poj2676。
代码如下:
#include <cstdio> #include <cstring> char s[10]; int a[10][10]; bool p; bool ok(int x,int y) { for(int i=0;i<4;i++) { if(i!=x && a[i][y]==a[x][y]) return false; } for(int i=0;i<4;i++) { if(i!=y && a[x][i]==a[x][y]) return false; } int nx=(x/2)*2; int ny=(y/2)*2; for(int i=nx;i<nx+2;i++) for(int j=ny;j<ny+2;j++) { if(i==x && j==y) continue; if(a[x][y]==a[i][j]) return false; } return true; } void dfs(int x,int y) { if(p || x==4) { p=true; return ; } if(a[x][y]) { if(y==3) dfs(x+1,0); else dfs(x,y+1); if(p) return; } else { for(int i=1;i<=4;i++) { a[x][y]=i; if(ok(x,y)) { // printf("%d\n",i); if(y==3) dfs(x+1,0); else dfs(x,y+1); } if(p) return; a[x][y]=0; } } } int main(void) { int n; scanf("%d",&n); for(int ca=1;ca<=n;ca++) { p=false; memset(a,0,sizeof(a)); for(int i=0;i<4;i++) { scanf("%s",s); for(int j=0;j<4;j++) { if(s[j]=='*') a[i][j]=0; else a[i][j]=s[j]-'0'; } } dfs(0,0); printf("Case #%d:\n",ca ); for(int i=0;i<4;i++) { for(int j=0;j<4;j++) printf("%d",a[i][j] ); printf("\n"); } } }
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