广度优先搜索

广度优先搜索小技巧

• 使用bool inq(in queue)来记录某一节点是不是入过队（而不是是否被访问）

矩阵的块 求给定的矩阵中块的个数

BFS解法

#include<queue>
#include<iostream>
using namespace std;
int X[] = { 0,0,1,-1 };
int Y[] = { 1,-1,0,0 };
const int max_x=30;
const int max_y=30;
bool inq[max_x][max_y] = { 0 };
int val[max_x][max_y];
struct point {
	int x;
	int y;
};
queue<point>q;
int x, y;
int cnt=0;
void BFS(int a,int b)
{
	point* t = new point;
	t->x = a;
	t->y = b;
	q.push(*t);
	inq[a][b] = true;
	point p;
	while (!q.empty())
	{
		p = q.front();
		q.pop();
		//添加上下左右
		if (val[p.x][p.y] == 1)
		{
			for (int i = 0; i < 4; i++)
			{
				int m = p.x + X[i], n = p.y + Y[i];
				if (m >= 0 && m < x &&
					n >= 0 && n < y)
				{
					if (!inq[m][n])
					{
						t = new point;
						t->x = m;
						t->y = n;
						q.push(*t);
						inq[m][n] = true;
					}
				}
			}
		}
	}
}
int main()
{
	scanf("%d%d", &x, &y);
	for (int i = 0; i < x; i++)
	{
		for (int j = 0; j < y; j++)
		{
			scanf("%d", &val[i][j]);
		}
	}
	for (int i = 0; i < x; i++)
	{
		for (int j = 0; j < y; j++)
		{
			if (!inq[i][j])
			{
				if (val[i][j] == 1)
				{
					cnt++;
					BFS(i,j);
				}
			}
		}
	}
	printf("%d", cnt);
}

DFS解法

//求给定的矩阵中块的个数
#include<queue>
#include<iostream>
using namespace std;
int X[] = { 0,0,1,-1 };
int Y[] = { 1,-1,0,0 };
const int max_x=30;
const int max_y=30;
bool isvisited[max_x][max_y] = { 0 };
int val[max_x][max_y];
struct point {
	int x;
	int y;
};
queue<point>q;
int x, y;
int cnt=0;
void DFS(int a,int b)
{
	isvisited[a][b] = true;
	//如果上下左右是1，访问
	for (int i = 0; i < 4; i++)
	{
		int m = a + X[i], n = b + Y[i];
		if (m >= 0 && m < x &&
			n >= 0 && n < y)
		{
			if (val[m][n] == 1)
			{
				if (!isvisited[m][n])
					DFS(m, n);
			}
		}
	}
}
int main()
{
	scanf("%d%d", &x, &y);
	for (int i = 0; i < x; i++)
	{
		for (int j = 0; j < y; j++)
		{
			scanf("%d", &val[i][j]);
		}
	}
	for (int i = 0; i < x; i++)
	{
		for (int j = 0; j < y; j++)
		{
			if (!isvisited[i][j])
			{
				if (val[i][j] == 1)
				{
					cnt++;
					DFS(i, j);
				}
			}
		}
	}
	printf("%d", cnt);
}