Codeforces Beta Round 32 (Div. 2, Codeforces format) D. Constellation 题解 枚举

Constellation

题目描述

A star map in Berland is a checked field $n\times m$ squares. In each square there is or there is not a star. The favorite constellation of all Berland’s astronomers is the constellation of the Cross. This constellation can be formed by any 5 stars so, that for some integer $x$ (radius of the constellation) the following is true:

• the 2nd is on the same vertical line as the 1st, but $x$ squares up
• the 3rd is on the same vertical line as the 1st, but $x$ squares down
• the 4th is on the same horizontal line as the 1st, but $x$ squares left
• the 5th is on the same horizontal line as the 1st, but $x$ squares right

Such constellations can be very numerous, that’s why they are numbered with integers from 1 on the following principle: when two constellations are compared, the one with a smaller radius gets a smaller index; if their radii are equal — the one, whose central star if higher than the central star of the other one; if their central stars are at the same level — the one, whose central star is to the left of the central star of the other one.

Your task is to find the constellation with index $k$ by the given Berland’s star map.

输入格式

The first line contains three integers $n$ , $m$ and $k$ ( $1<=n,m<=300,1<=k<=3\times 10^7$ ) — height and width of the map and index of the required constellation respectively. The upper-left corner has coordinates $(1,1)$ , and the lower-right — $(n,m)$ . Then there follow $n$ lines, $m$ characters each — description of the map. $j$ -th character in $i$ -th line is “*”, if there is a star in the corresponding square, and “.” if this square is empty.

输出格式

If the number of the constellations is less than $k$ , output -1. Otherwise output 5 lines, two integers each — coordinates of the required constellation. Output the stars in the following order: central, upper, lower, left, right.

题面翻译

题目描述

一个Berland星空图填充了一个 $N\times M$ 的矩形。Berland的十字星座是所有的天文学家最喜欢的星座。这个星座由 $5$ 个星星组成。对于星座的半径 $x$，$5$ 个星星的位置关系有以下原则：

第二个和第一个在同一条垂直线上，但是在第一个的上边 $x$ 单位处。

第三个和第一个在同一条垂直线上，但是在第一个的下边 $x$ 单位处。

第四个和第一个在同一水平线上，但是在第一个的左边 $x$ 单位处。

第五个和第一个在同一水平线上，但是在第一个的右边 $x$ 单位处。

对于星座的索引：更小的半径会有更小的索引。如果它们的半径相等，位于更上边更左边的中心恒星所在的十字星座索引更小。

你的任务是找到索引为 $K$ 的星座的Berland星空图。

输入格式

第一行包含三个整数 $N,M$ 和 $K$($1\le N,M\le 300，1\le K\le 3\times 10^7$)，分别为地图的高，宽和所求星座的索引。下面输入 $N$ 行，$M$ 列的字符来描述星空图。左上角坐标为 $(1,1)$，右下角坐标为 $(N,M)$。

输出格式

如果星座索引小于 $k$，输出 $-1$。否则输出 $5$ 行，每行分别两个整数，即星座的每一个星星的坐标。按照中心、上、下、左、右的顺序输出星星。

样例 #1

样例输入 #1

5 6 1
....*.
...***
....*.
..*...
.***..

样例输出 #1

2 5
1 5
3 5
2 4
2 6

样例 #2

样例输入 #2

5 6 2
....*.
...***
....*.
..*...
.***..

样例输出 #2

-1

样例 #3

样例输入 #3

7 7 2
...*...
.......
...*...
*.***.*
...*...
.......
...*...

样例输出 #3

4 4
1 4
7 4
4 1
4 7

原题

Codeforces——传送门
洛谷——传送门

代码

#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
typedef long long ll;

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);

    int n, m, k;
    cin >> n >> m >> k;
    vector<vector<char>> g(n + 1, vector<char>(m + 1, 0)); // 存图
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            cin >> g[i][j];
        }
    }
    int cnt = 0;                                             // 当前十字星座个数
    for (int r = 1; r <= max((n - 1) / 2, (m - 1) / 2); r++) // 枚举半径
    {
        // 枚举中心星的坐标
        for (int i = 1 + r; i <= n - r; i++)
        {
            for (int j = 1 + r; j <= m - r; j++)
            {
                // 十字星座的五个顶点均为'*'
                if (g[i][j] == '*' && g[i - r][j] == '*' && g[i + r][j] == '*' && g[i][j - r] == '*' && g[i][j + r] == '*')
                {
                    cnt++;        // 星座个数++
                    if (cnt == k) // 找到第k个星座，直接输出答案
                    {
                        cout << i << ' ' << j << '\n';
                        cout << i - r << ' ' << j << '\n';
                        cout << i + r << ' ' << j << '\n';
                        cout << i << ' ' << j - r << '\n';
                        cout << i << ' ' << j + r << '\n';
                        return 0;
                    }
                }
            }
        }
    }
    // 星座数小于k，输出-1
    cout << -1 << '\n';

    return 0;
}