Uva 11520 - Fill the Square 贪心 难度: 0

Posted 2021-02-09 xuesu

题目

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2515

题意

n*n矩阵中填入大写字母，n <= 10，要求从上到下从左到右字母序最小，相邻格子字母不同

 

思路

如刘书思路，状态比较小，不会导致矛盾。

 

感想

1. 状态较小

 

代码

#include <algorithm>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <tuple>
#define LOCAL_DEBUG
using namespace std;
typedef pair<double, int> MyPair;
const int MAXN = 11;
int dx[4] = { 1, -1, 0, 0 };
int dy[4] = { 0, 0, 1, -1 };
char maze[MAXN][MAXN];
bool used[26];
void fillin(int i, int j, int n) {
    memset(used, 0, sizeof used);
    for (int di = 0; di < 4; di++) {
        int tx = i + dx[di];
        int ty = j + dy[di];
        if (0 > tx || tx >= n)continue;
        if (0 > ty || ty >= n)continue;
        if (maze[tx][ty] == ‘.‘)continue;
        used[maze[tx][ty] - ‘A‘] = true;
    }
}


int main() {
#ifdef LOCAL_DEBUG
    freopen("C:\Users\Iris\source\repos\ACM\ACM\input.txt", "r", stdin);
    //freopen("C:\Users\Iris\source\repos\ACM\ACM\output.txt", "w", stdout);
#endif // LOCAL_DEBUG
    int T;
    scanf("%d", &T);
    for (int ti = 1; ti <= T; ti++) {
        int n;
        scanf("%d", &n);
        for (int i = 0; i < n; i++)scanf("%s", maze + i);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (maze[i][j] != ‘.‘)continue;
                fillin(i, j, n);
                for (char c = ‘A‘; c <= ‘Z‘; c++) {
                    if (!used[c - ‘A‘]) {
                        maze[i][j] = c;
                        break;
                    }
                }
            }
        }
        printf("Case %d:
", ti);
        for (int i = 0; i < n; i++)printf("%s
", maze[i]);
        
    }

    return 0;
}

View Code