UVA 11520 - Fill the Square (贪心)

最新推荐文章于 2023-04-18 12:51:05 发布

最新推荐文章于 2023-04-18 12:51:05 发布 · 72 阅读

本文介绍了一个填字游戏的实现算法，旨在确保填充后的网格字典序最小。文章详细阐述了如何在遵循特定规则的情况下填充网格，并提供了一段C语言代码示例。

Problem A
Fill the Square
Input:Standard Input

Output:Standard Output

In this problem, you have to draw a square using uppercase English Alphabets.

To be more precise, you will be given a square grid with some empty blocks and others already filled for you with some letters to make your task easier. You have to insert characters in every empty cell so that the whole grid is filled with alphabets. In doing so you have to meet the following rules:

Make sure no adjacent cells contain the same letter; two cells are adjacent if they share a common edge.
There could be many ways to fill the grid. You have to ensure you make the lexicographically smallest one. Here, two grids are checked in row major order when comparing lexicographically.

Input
The first line of input will contain an integer that will determine the number of test cases. Each case starts with an integern( n<=10 ),that represents the dimension of the grid. The nextnlines will containncharacters each. Every cell of the grid is either a ‘.’ or a letter from [A, Z]. Here a ‘.’ Represents an empty cell.

Output
For each case, first output Case #: ( # replaced by case number ) and in the next n lines output the input matrix with the empty cells filled heeding the rules above.

Sample Input Output for Sample Input

...

A..

...

Case 1:

ABA

BAB

ABA

Case 2:

BAB

ABA

BAB

题意：给定一个map，要填充使得字典序最小。求填充之后样子。

思路:每个去填充的时候，都找个最小的去填充即可。

代码：

#include <stdio.h>
#include <string.h>
#define max(a,b) (a)>(b)?(a):(b)
const int d[4][2] = {{-1, 0}, {0, -1}, {1, 0}, {0, 1}};
const int N = 15;

int t, n;
char g[N][N];

void init() {
    scanf("%d", &n);
    for (int i = 0; i < n; i ++)
	scanf("%s", g[i]);
}

bool judge(int x, int y) {
    if (x >= 0 && x < n && y >= 0 && y < n)
	return true;
    return false;
}

bool judge2(int i, int x, int y) {
    for (int j = 0; j < 4; j ++) {
	int xx = x + d[j][0];
	int yy = y + d[j][1];
	if (judge(xx, yy)) {
	    if (i + 'A' == g[xx][yy])
		return false;
	}
    }
    return true;
}
char find(int x, int y) {
    for (int i = 0; i < 26; i ++) {
	if (judge2(i, x, y))
	    return i + 'A';
    }
}

void solve() {
    for (int i = 0; i < n; i ++) {
	for (int j = 0; j < n; j ++) {
	    if (g[i][j] == '.')
		g[i][j] = find(i, j);
	    printf("%c", g[i][j]);
	}
	printf("\n");
    }
}

int main() {
    int cas = 0;
    scanf("%d", &t);
    while (t--) {
	init();
	printf("Case %d:\n", ++cas);
	solve();
    }
    return 0;
}