Sunday, June 12, 2016

[hihoCoder] 岛屿 解题报告 - Jonah的专栏 - 博客频道 - CSDN.NET


[hihoCoder] 岛屿 解题报告 - Jonah的专栏 - 博客频道 - CSDN.NET
给你一张某一海域卫星照片,你需要统计:
1. 照片中海岛的数目
2. 照片中面积不同的海岛数目
3. 照片中形状不同的海盗数目

其中海域的照片如下,"."表示海洋,"#"表示陆地。在"上下左右"四个方向上连在一起的一片陆地组成一座岛屿。
.####..  
.....#.  
####.#.  
.....#.  
..##.#.  

输入

第一行包含两个人整数:N 和 M,(1 ≤ NM ≤ 50),表示照片的行数和列数。
以下一个 N * M 的矩阵,表示表示海域的照片。

输出

输出3个整数,依次是照片中海岛的数目、面积不同的海岛数目和形状不同的海岛数目。
上图所示的照片中一共有4座岛屿;其中3座面积为4,一座面积为2,所以不同面积的岛屿数目是2;有两座形状都是"####",所以形状不同的岛屿数目为3。
样例输入
5 7
.####..  
.....#.  
####.#.  
.....#.  
..##.#.  
样例输出
4 2 3

思路: 一个普通的DFS的延伸, 其中岛屿的个数比较容易计算, 就是遍历数组碰到'#'就进行DFS, 并且为了防止再次搜索到这个点, 我们可以搜过之后改变其值. 面积就是每次搜索到的'#'的个数, 也比较容易. 海岛形状这个我们可以保存搜到的海岛的位置, 并且以最初的起点的岛屿为相对值0, 一个位置可以表示成(y*n + x), 这样我们就可以以一个数的形式保存一个位置, 保存的时候都减去起始点的值, 然后将其保存的二叉搜索树中去, 这样可以保持其大小有序. 然后看其不同的个数有多少即可.
int M, N, num = 0;
vector<vector<char> > map;
set<int> area;
set<set<int> > shape;

void DFS(pair<int, int> curPos, pair<int, int> oriPos, set<int>& st)
{
    int y=curPos.first, x=curPos.second, oy=oriPos.first, ox=oriPos.second;
    if(y>=N || y<0 || x<0 || x>=M || map[y][x]!='#') return; 
    st.insert(y*M+x - oy*M-ox);
    map[y][x] = '.';
    DFS(make_pair(y+1, x),oriPos, st);
    DFS(make_pair(y-1, x),oriPos, st);
    DFS(make_pair(y, x+1),oriPos, st);
    DFS(make_pair(y, x-1),oriPos, st);
}

int main()
{
    cin >> N >> M;
    for(int i = 0; i < N; i++)
    {
        vector<char> vec;
        char ch;
        for(int j =0; j < M; j++)
        {
            cin>> ch;
            vec.push_back(ch);
        }
        map.push_back(vec);
    }
    for(int i = 0; i < N; i++)
        for(int j =0; j < M; j++)
            if(map[i][j] == '#')
            {
                set<int> st{0};
                DFS(make_pair(i, j),make_pair(i, j), st);
                area.insert((int)st.size());   
                shape.insert(st);
                num++;
            }    
    cout << num << " " << area.size() << " " << shape.size() << endl;
    return 0;
}
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