POJ 3026 Borg Maze BFS+Prim

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本文介绍了一种使用BFS寻找'S'与每个'A'之间的最短路径，然后通过Prim算法构建迷宫内连接所有点的最小生成树的方法。详细解释了算法流程并提供了代码实现。

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题意：给定一个迷宫，在一个迷宫内，建立一颗最小生成树连接所有点。（这些点即‘A’或‘S’)
题解：通过BFS找到'S'与每个’A'之间的最短路径。然后prim 建立最小生成树。


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

#define INF 10000000
#define N 52

bool vis[N][N];
int path[N*2][N*2], map[N][N];
int dir[4][2]={{-1,0},{0,1},{1,0},{0,-1}};
int row, col, cnt, Brog;

struct Node
{
	int x, y, step;
}   node[N*2];

bool isOk ( int a, int b )
{
	if ( a < 0 || a > row || b < 0 || b > col || vis[a][b] == 1 || map[a][b] == -1 )
		return false;
	return true;
}

void bfs ()
{
	Node current, next;
	queue<Node> Q;
	memset(path,0,sizeof(path));

	for ( int i = 1; i <= Brog; i++ )
	{
		while ( ! Q.empty() ) Q.pop();
		memset(vis,0,sizeof(vis));
		node[i].step = cnt = 0;
		vis[node[i].x][node[i].y] = 1;
		Q.push ( node[i] );   
		
		while ( ! Q.empty () )
		{
			current = Q.front ();
			Q.pop ();
			int x = current.x;
			int y = current.y;

			if ( map[x][y] != 0 && map[x][y] != -1 )
			{
				cnt++;
				path[i][map[x][y]] = path[map[x][y]][i] = current.step;
				if ( cnt == Brog ) break;
			}

			for ( int j = 0; j < 4; j++ )
			{
				x = current.x + dir[j][0];
				y = current.y + dir[j][1];
				if ( isOk ( x, y ) )
				{
					next.x = x;
					next.y = y;
					next.step = current.step + 1;
					vis[x][y] = 1;
					Q.push ( next );
				}
			}
		}
	}
}


int prim()
{
	int i, j, k, min, sum = 0;
	int dis[102], mark[102] = { 0 };

    for ( i = 1; i <= Brog; i++ )
		dis[i] = INF;
	dis[1] = 0;
	for ( i = 1; i <= Brog; i++ )
	{
		min = INF;
		for ( j = 1; j <= Brog; j++ )
		{
			if ( ! mark[j] && dis[j] < min )
			{
				k = j;
				min = dis[j];
			}
		}
		sum += min;
		mark[k] = 1;
		for ( j = 1; j <= Brog; j++ )
		{
			if ( ! mark[j] && path[k][j] < dis[j] )
				dis[j] = path[k][j];
		}
	}
	return sum;
}


int main()
{
	int t;
	char str[52];
	cin >> t;
	while ( t-- )
	{
		cin >> col >> row;
		cin.getline (str,52); /* 输入需要注意 */

		Brog = 0;
		memset(node,0,sizeof(node));

		for ( int i = 0; i < row; i++ )
		{
			cin.getline(str, 52, '\n');
			for ( int j = 0; j < col; j++ )
			{
				if ( str[j] == 'A' || str[j] == 'S' )
				{
					Brog++;
					map[i][j] = Brog;
					node[Brog].x = i;
					node[Brog].y = j;   
				}
				else if ( str[j] == ' ' )
					map[i][j] = 0;
				else if ( str[j] == '#' )
					map[i][j] = -1;
			}
		}
		
		bfs();
		cout << prim() << endl;
	}
	return 0;
}