UVA - 11520 - Fill the Square（贪心）

Posted 2020-08-15 TianTengtt

题意：给定一个n * n（1 <= n <= 10）的网格，有些已经填上了一些大写字母，需要补充完所有的字母，使每两两相邻的格子中的字母不同，且从上至下，从左至右，组成一个字符串后字典序最小。

由于组成字符串后的长度都为n * n，故字典序越往前的字母决定的优先级越大，所以贪心即可，从上到下，从左到右的补充格子里的字母，尽可能的使当前可以补充的字母尽量小。

 

#include<cstdio>  
#include<cstring>  
#include<cctype>  
#include<cstdlib>  
#include<cmath>  
#include<iostream>  
#include<sstream>  
#include<iterator>  
#include<algorithm>  
#include<string>  
#include<vector>  
#include<set>  
#include<map>  
#include<deque>  
#include<queue>  
#include<stack>  
#include<list>  
#define fin freopen("in.txt", "r", stdin)  
#define fout freopen("out.txt", "w", stdout)  
#define pr(x) cout << #x << " : " << x << "   "  
#define prln(x) cout << #x << " : " << x << endl  
typedef long long ll;  
typedef unsigned long long llu;  
const int INT_INF = 0x3f3f3f3f;  
const int INT_M_INF = 0x7f7f7f7f;  
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;  
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;  
const double pi = acos(-1.0);  
const double EPS = 1e-6;  
const int dx[] = {0, 0, -1, 1};  
const int dy[] = {-1, 1, 0, 0};  
const ll MOD = 1e9 + 7;  
const int MAXN = 10 + 10;  
const int MAXT = 10000 + 10;  
using namespace std;  
  
int T, n;  
char tu[MAXN][MAXN];  
  
int main(){  
    int kase = 0;  
    scanf("%d", &T);  
    while(T--){  
        scanf("%d", &n);  
        for(int i = 0; i < n; ++i)  scanf("%s", tu[i]);  
        for(int i = 0; i < n; ++i)  
        for(int j = 0; j < n; ++j){  
            if(tu[i][j] != ‘.‘)  continue;  
            for(char ch = ‘A‘; ch <= ‘Z‘; ++ch){  
                bool flag = true;  
                for(int k = 0; k < 4; ++k)  
                    if(i + dx[k] >= 0 && i + dx[k] < n && j + dy[k] >= 0 && j + dy[k] <= n && tu[i + dx[k]][j + dy[k]] == ch){  
                        flag = false;  
                        break;  
                    }  
                if(flag){  
                    tu[i][j] = ch;  
                    break;  
                }  
            }  
        }  
        printf("Case %d:\n", ++kase);  
        for(int i = 0; i < n; ++i)  printf("%s\n", tu[i]);  
    }  
    return 0;  
}