Codeforces_389B_Fox and Cross(贪心)

最新推荐文章于 2020-10-06 19:46:00 发布

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#Codeforces #贪心

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本文解析了一道Codeforces竞赛题目，该题要求玩家在由‘#’和‘.’构成的n*n的地图中，判断所有‘#’是否能构成不重叠的十字图案。通过直接遍历地图并应用贪心算法解决此问题。

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Fox Ciel has a board with n rows and n columns. So, the board consists of n × n cells. Each cell contains either a symbol '.', or a symbol '#'.

A cross on the board is a connected set of exactly five cells of the board that looks like a cross. The picture below shows how it looks.

$\text{[math]}$

Ciel wants to draw several (may be zero) crosses on the board. Each cross must cover exactly five cells with symbols '#', and any cell with symbol '#' must belong to some cross. No two crosses can share a cell.

Please, tell Ciel if she can draw the crosses in the described way.

Input

The first line contains an integer n (3 ≤ n ≤ 100) — the size of the board.

Each of the next n lines describes one row of the board. The i-th line describes the i-th row of the board and consists of n characters. Each character is either a symbol '.', or a symbol '#'.

Output

Output a single line with "YES" if Ciel can draw the crosses in the described way. Otherwise output a single line with "NO".

Sample test(s)

input
5
.#...
####.
.####
...#.
.....

output
YES

input
4
####
####
####
####

output
NO

input
6
.#....
####..
.####.
.#.##.
######
.#..#.

output
YES

input
6
.#..#.
######
.####.
.####.
######
.#..#.

output
NO

input
3
...
...
...

output
YES

Note

In example 1, you can draw two crosses. The picture below shows what they look like.

$\text{[math]}$

In example 2, the board contains 16 cells with '#', but each cross contains 5. Since 16 is not a multiple of 5, so it's impossible to cover all.

题型：贪心

题意：在由‘#’和‘.’构成的n*n的地图中，问所有的‘#’能否不重叠的构成多个十字，如果全是‘.’也输出YES

分析：

直接贪心遍历，如果当前的一个点是‘#‘并且其上下左右也是’#’，那么就将这五个点全部改成‘#’，最后再将地图遍历一遍，如果还存在‘#’，则输出NO。

代码：

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
using namespace std;
int n;
char map[123][123];

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

bool check(int x,int y) {
    if(isin(x-1,y) && isin(x+1,y) && isin(x,y+1) && isin(x,y-1)) {
        if(map[x-1][y]=='#' && map[x+1][y]=='#' && map[x][y+1]=='#' && map[x][y-1]=='#')
            return true;
    }
    return false;
}

void change(int x,int y){
    map[x][y] = '.';
}

int main() {
    while(~scanf("%d",&n)) {
        for(int i=0; i<n; i++) {
            scanf("%s",map[i]);
        }
        for(int i=0; i<n; i++) {
            for(int j=0; j<n; j++) {
                if(map[i][j]=='#' && check(i,j))
                {
                    change(i,j);
                    change(i-1,j);
                    change(i+1,j);
                    change(i,j-1);
                    change(i,j+1);
                }
            }
        }
        bool flag = true;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(map[i][j] == '#'){
                    flag = false;
                    break;
                }
            }
        }
        if(flag) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}