HDU 5547 Sudoku (暴力)

题意：数独。

析：由于只是4*4，完全可以暴力，要注意一下一些条件，比如2*2的小方格也得是1234

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 10 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
char s[maxn][maxn];
int a[maxn][maxn];

bool judge(int r, int c, int x){
  int cnt = 0;
  for(int i = 0; i < 4; ++i)
    if(a[r][i] == x) ++cnt;
  if(cnt > 0)  return false;
  cnt = 0;
  for(int i = 0; i < 4; ++i)
    if(a[i][c] == x)  ++cnt;
  return cnt < 1;
}

bool solve(){
  set<int> sets;
  for(int i = 0; i < 2; ++i)
    for(int j = 0; j < 2; ++j)
      sets.insert(a[i][j]);
  if(sets.size() != 4)  return false;
  sets.clear();
  for(int i = 2; i < 4; ++i)
    for(int j = 2; j < 4; ++j)
      sets.insert(a[i][j]);
  if(sets.size() != 4)  return false;
  sets.clear();
  for(int i = 2; i < 4; ++i)
    for(int j = 0; j < 2; ++j)
      sets.insert(a[i][j]);
  if(sets.size() != 4)  return false;
  sets.clear();
  for(int i = 0; i < 2; ++i)
    for(int j = 2; j < 4; ++j)
      sets.insert(a[i][j]);
  if(sets.size() != 4)  return false;
  return true;
}

bool dfs(int r, int c){
  if(a[r][c]){
    if(r == 3 && c == 3)  return solve();
    return c == 3 ? dfs(r+1, 0) : dfs(r, c+1);
  }
  for(int i = 1; i < 5; ++i)  if(judge(r, c, i)){
    a[r][c] = i;
    int rr = r, cc = c;
    if(r == 3 && c == 3){
      if(solve())  return true;
      a[r][c] = 0;
      continue;
    }
    if(c == 3)  ++rr, cc = 0;
    else ++cc;
    if(dfs(rr, cc))  return true;
    a[r][c] = 0;
  }
  return false;
}

int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    for(int i = 0; i < 4; ++i)  scanf("%s", s+i);
    memset(a, 0, sizeof a);
    for(int i = 0; i < 4; ++i)
      for(int j = 0; j < 4; ++j)
        if(s[i][j] != '*')  a[i][j] = s[i][j] - '0';
    dfs(0, 0);
    printf("Case #%d:\n", kase);
    for(int i = 0; i < 4; ++i, printf("\n"))
      for(int j = 0; j < 4; ++j)
        printf("%d", a[i][j]);
  }
  return 0;
}