文章目录

• 题目大意
• 解题思路
• 代码

题目大意

三维迷宫中有墙(X)，路(.)，起点(S)，终点(E), 问终点是否可达，若可达最早什么时候可达。

解题思路

• (1) 最短时间到达，直接BFS。
• (2) 注意grad[z][x][y]，第一维代表纵坐标。
• (3) BFS用优先队列较好。

代码

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
const int MAX = 35;
struct Node
{
    int x,y,z,t;
    bool operator < (const Node &A)const{
        return t > A.t;
    }
}tmp;

priority_queue<Node, vector<Node> >PQ;
char grad[MAX][MAX][MAX];
int visited[MAX][MAX][MAX];
int dir[6][3] = {{1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1}};
int L, n, m;
int bfs(int sx, int sy, int sz)
{
    while(!PQ.empty())
        PQ.pop();
    memset(visited, 0, sizeof(visited));
    tmp.x = sx;
    tmp.y = sy;
    tmp.z = sz;
    tmp.t = 0;
    PQ.push(tmp);
    while(!PQ.empty())
    {

        Node top = PQ.top();
        PQ.pop();
        //cout << top.x << " " << top.y << " " << top.z << endl;
        visited[top.z][top.x][top.y] = 1;
        for(int i=0; i<6; i++)
        {
            int nx = top.x + dir[i][0];
            int ny = top.y + dir[i][1];
            int nz = top.z + dir[i][2];

            if(nx < 0 || nx >=n || ny < 0 || ny >= m || nz < 0 || nz >= L)
                continue;
            if(visited[nz][nx][ny])
                continue;
            if(grad[nz][nx][ny] == '#')
                continue;
            if(grad[nz][nx][ny] == 'E')
                return top.t + 1;

            tmp.x = nx;
            tmp.y = ny;
            tmp.z = nz;
            tmp.t = top.t+1;
            visited[nz][nx][ny] = 1;
            PQ.push(tmp);
        }
    }
    return -1;

}
int main()
{
    int sx, sy, sz;
    while(cin >> L >> n >> m)
    {
        if(L == 0 && n == 0 && m == 0)
            break;
        for(int i=0; i<L; i++)
        {
            for(int j=0; j<n; j++)
            {
                for(int k=0; k<m; k++)
                {
                    cin >> grad[i][j][k];
                    if(grad[i][j][k] == 'S')
                    {
                        sx = j;
                        sy = k;
                        sz = i;
                    }
                }
            }
        }
        int ans = bfs(sx, sy, sz);
        if(ans == -1)
            cout << "Trapped!" << endl;
        else
            cout << "Escaped in "<<ans<<" minute(s)." << endl;
    }
    return 0;
}

标签：tmp,Dungeon,PQ,NOI,int,MAX,ny,nz,Master
来源： https://blog.csdn.net/Wangpeiyi9979/article/details/94385654