UVa 11520 Fill the Square (水题，暴力)

题意：给n*n的格子里填上A-Z的字符，保证相邻字符不同，并且字典序最小。

析：直接从第一个格子开始暴力即可，每次判断上下左是不是相同即可。

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 <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
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 = 1e4 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -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[15][15];
bool ok;

bool dfs(int r, int c){
    if(isalpha(s[r][c])){
        if(r == n-1 && c == n-1)  return true;
        else if(c == n-1)  return dfs(r+1, 0);
        else return dfs(r, c+1);
    }
    for(int i = 0; i < 26; ++i){
        if(r > 0 && s[r-1][c] == 'A' + i)  continue;
        if(c > 0 && s[r][c-1] == 'A' + i)  continue;
        if(r < n-1 && s[r+1][c] == 'A' + i)  continue;
        if(c < n-1 && s[r][c+1] == 'A' + i)  continue;
        s[r][c] = 'A' + i;
        if(r == n-1 && c == n-1)  return true;
        else if(c == n-1)  return dfs(r+1, 0);
        else return dfs(r, c+1);
    }
    return true;
}

int main(){
    int T;  cin >> T;
    for(int kase = 1; kase <= T; ++kase){
        scanf("%d", &n);
        for(int i = 0; i < n; ++i)  scanf("%s", s+i);
        ok = false;
        dfs(0, 0);
        printf("Case %d:\n", kase);
        for(int i = 0; i < n; ++i)  puts(s[i]);
    }
    return 0;
}