FloodFill 统计子图形的两种方式

最新推荐文章于 2018-11-13 15:00:28 发布

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#Flood Fill

Flood Fill算法是图论中非常重要的一种算法思想：找到一个符合要求的点，按照某种方式对该点进行扩充，标记该点，对扩充的点进行重复操作。

Flood Fill的两种典型的实现方式是bfs和dfs，下面两道统计子图形的题目，分别用bfs和dfs实现

1. http://222.200.185.45/7148

描述：给出图形，问其中有多少个矩形，多少个圆形，多少个三角形。

注意题目要求：每个图形至少包含15个‘#’，空缺最多占10%，任意两个图形不共边，不共点（这点非常重要，也就是说每一个图形都是完全孤立的）

思路：用bfs进行Flood Fill，对于每个图形，统计其中‘#’的个数，算出包含该图形的最小矩形的大小，根据图形占矩形的百分比判断该图形的形状。

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

struct Point
{
	int x, y;
	Point() {}
	Point(int _x, int _y): x(_x), y(_y) {}
};
int r, c, re, tr, ci;
char data[510][510];
int dir[8][2] = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
void floodfill(int x, int y)
{
	queue <Point> Q;
	Q.push(Point(x,y));
	data[x][y] = '.';
	int cnt = 1;
	int minx = x, maxx = x, miny = y, maxy = y;
	Point cur, nxt;
	while (!Q.empty())
	{
		cur = Q.front();
		Q.pop();
		for (int i = 0; i < 8; ++ i)
		{
			nxt = Point(cur.x+dir[i][0], cur.y+dir[i][1]);
			if (data[nxt.x][nxt.y] == '#')
			{
				Q.push(nxt);
				data[nxt.x][nxt.y] = '.';
				cnt += 1;
				if (minx > nxt.x) minx = nxt.x;
				if (maxx < nxt.x) maxx = nxt.x;
				if (miny > nxt.y) miny = nxt.y;
				if (maxy < nxt.y) maxy = nxt.y;
			}
		}
	}
	if (cnt < 15) return;
	double area = 0.0 + (maxx-minx+1)*(maxy-miny+1);
	if (cnt / area > 0.875) re += 1;
	else if (cnt / area > 0.6) ci += 1;
	else if (cnt / area > 0.1) tr += 1;
}
int main()
{
	while (~scanf("%d %d", &r, &c))
	{
		re = tr = ci = 0;
		for (int i = 1; i <= r; ++ i)
		{
			scanf("%s", data[i]+1);
			data[i][c+1] = '.';
		}
		for (int i = 1; i <= r; ++ i)
			for (int j = 1; j <= c; ++ j)
				if (data[i][j] == '#')
					floodfill(i, j);
		printf("Rectangle: %d\nCircle: %d\nTriangle: %d\n\n", re, ci, tr);
	}
}

1008. Triangle, rectangle and square

Time Limit: 1sec Memory Limit:256MB

Description

We represent a binary image with a 2D array of bits. The bits consist of only 0 and 1. The eight surrounding bits around a bit are its connected neighbors. The image has several shapes that consist of connected 1's. The shapes can be triangles, rectangles, squares. Now given an image, you are to count the number of each kind of shape. If a shape is a square, don't count it as a rectangle. All 1's on a side of any shape lie on a straight line. We ensure that no any two shapes are connected, and any given shape contains 0's inside it.

Input

The input file will consist of several test cases.

The first line of each case is two positive integers M and N (3<=M, N<=100), which means that the image grid has M rows and N columns. Then M lines follow, each line with a string of N characters. The characters would only be "0" or "1".

The input is ended with M=0 and N=0.

Output

For each test case, output on a single line the number of each kind of shape, with the format "Triangle: A, Rectangle: B, Square: C". A, B and C are respectively the numbers of triangle, rectangle and square.

Sample Input

Sample Output

Triangle: 1, Rectangle: 1, Square: 0
Triangle: 0, Rectangle: 0, Square: 1

Problem Source: wuhefeng

描述：给出图形，问其中有多少个三角形，长方形，正方形。这次是包围式的。

注意题目要求：任意两个图形不共边，不共点（这点非常重要，也就是说每一个图形都是完全孤立的）。每个图形中间用空白填充。

思路：用dfs进行Flood Fill，对于每个图形，记录它的顶点，根据顶点的个数判断是三角形还是矩形，如果是矩形，再根据图形的边长判断是长方形还是正方形。

#include <cstdio>
#include <vector>
#include <cmath>
#include <cstring>
using namespace std;
typedef pair<int,int> pii;
int r, c, tr, re, sq;
char data[110][110];
int dir[8][2] = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
vector <pii> point;
void floodfill(int x, int y)
{
	point.push_back(pii(x,y));
	int tx, ty;
	for (int i = 0; i < 8; ++ i)
	{
		tx = x + dir[i][0];
		ty = y + dir[i][1];
		if (data[tx][ty] == '1')
		{
			data[tx][ty] = '0';
			while (data[tx+dir[i][0]][ty+dir[i][1]] == '1')
			{
				tx += dir[i][0];
				ty += dir[i][1];
				data[tx][ty] = '0';
			}
			floodfill(tx,ty);
		}
	}
}
void judge()
{
	if (point.size() == 4) tr += 1;
	else if (point.size() == 5)
	{
		if (abs(point[0].first-point[2].first) == abs(point[0].second-point[2].second))
			sq += 1;
		else if (point[0].first == point[2].first || point[0].second == point[2].second)
			sq += 1;
		else re += 1;
	}
}
int main()
{
	while (true)
	{
		tr = re = sq = 0;
		memset(data, '0', sizeof(data));
		scanf("%d %d", &r, &c);
		if (r == 0 && c == 0) break;
		for (int i = 1; i <= r; ++ i)
		{
			scanf("%s", data[i]+1);
			data[i][c+1] = '0';
		}
		for (int i = 1; i <= r; ++ i)
		{
			for (int j = 1; j <= c; ++ j)
			{
				if (data[i][j] == '1')
				{
					point.clear();
					floodfill(i, j);
					judge();
				}
			}
		}
		printf("Triangle: %d, Rectangle: %d, Square: %d\n", tr, re, sq);
	}
}