hihocoder1310 岛屿

hihocoder1310 岛屿

题意：

中文题意

思路：

dfs,面积和数量都很好求，问题在岛屿形状上，感觉让人比较麻烦，用vector保存各个点，只要两个岛之间每个点距离一样就好了，这里的形状的定义比较狭隘，就是平移可得的意思，如果换成可以旋转等变换得到，会比较麻烦感觉。

ac代码：

C++

// daoyu1310.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"

#include<cstdio>
#include<iostream>
#include<string>
#include<set>
#include<vector>

using namespace std;

int n, m;			//1 <= n, m <= 50
char map[51][51];			//store map		
int dir[4][2] = { 1,0,-1,0,0,1,0,-1 };

struct island {
	int area;
	vector<pair<int, int> > shape;
	island(int x, int y, int s)
	{
		shape.push_back(make_pair(x, y));
		area = s;
	}
};

bool issame(island* a, island* b)
{
	if (a->area != b->area) return false;
	for (int k = 1; k < a->shape.size(); k++)
	{
		if (a->shape[k] != b->shape[k]) return false;
	}
	return true;
}

int count(int i, int j, island* is)
{
	if (map[i][j] == '#')
	{
		map[i][j] = '*';			//represent visited
		is->shape.push_back(make_pair(i - is->shape[0].first, j - is->shape[0].second));
	}
	else
	{
		return 0;
	}

	int s = 1;
	int nexti, nextj;
	for (int k = 0; k < 4; k++)
	{
		nexti = i + dir[k][0];
		nextj = j + dir[k][1];
		if (nexti >= 0 && nexti < n && nextj >= 0 && nextj < m && map[nexti][nextj] == '#')
			s += count(nexti, nextj, is);
	}
	return s;

}

int main()
{
	cin >> n >> m;
	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++)
			cin >> map[i][j];

	vector<island*> res;
	set<int> area;
	int countshape = 0;
	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++)
		{
			if (map[i][j] == '#')
			{
				island* il = new island(i, j, 1);
				il->area = count(i, j, il);
				area.insert(il->area);
				countshape++;
				for (int k = 0; k < res.size(); k++)
				{
					if (issame(il, res[k]))
					{
						countshape--;
						break;
					}
				}
				res.push_back(il);
			}
		}
	//for (int i = 0; i < res.size(); i++)
	//{
	//	cout << res[i]->area << endl;
	//	for (int j = 0; j < res[i]->shape.size(); j++)
	//	{			
	//		cout << res[i]->shape[j].first << " " << res[i]->shape[j].second<< endl;
	//	}
	//	cout << endl;
	//	
	//}
	//if (issame(res[0],res[2])) cout << "hello" << endl;
	cout << res.size() << " " << area.size() << " " << countshape << endl;
    return 0;
}