poj 2251 三维地图最短路径问题 bfs算法

题意:给你一个三维地图,然后让你走出去,找到最短路径。

思路:bfs

  1. 每个坐标的表示为 x,y,z并且每个点都需要加上时间 t  

    struct node
    {
    int x, y, z;
    int t;
    };

  2. bfs用队列,进队列的时候要标记,并且 t+1;
  3. 最先到达终点的,所花的时间必定最短

代码上的小技巧:
三维地图需要你去遍历的时候需要走六个方向:

int dx[6] = { 0,0,0,0,1,-1 };
int dy[6] = { 0,0,-1,1,0,0 };
int dz[6] = { 1,-1,0,0,0,0 };

解决问题的代码:

#include <cstdio>
#include <iostream>
#include <queue>
#include <algorithm>
#include <cstring>
using namespace std;
int r, n, m;
string s[35][35];
struct node
{
    int x, y, z;
    int t;
};
node Begin, End;
queue <node> que;
int dx[6] = { 0,0,0,0,1,-1 };
int dy[6] = { 0,0,-1,1,0,0 };
int dz[6] = { 1,-1,0,0,0,0 };
int vis[35][35][35];
int bfs()
{
    que.push(Begin);
    vis[Begin.x][Begin.y][Begin.z] = 1;
    while (!que.empty())
    {
        node front = que.front();
        que.pop();
        if (front.x == End.x && front.y == End.y && front.z == End.z) return front.t;
        for (int i = 0; i < 6; i++)
        {
            node tmp;
            tmp.x = front.x + dx[i];
            tmp.y = front.y + dy[i];
            tmp.z = front.z + dz[i];
            tmp.t = front.t + 1;
            if (tmp.x >= 0 && tmp.x < r && tmp.y >= 0 && tmp.y < n && tmp.z >= 0 && tmp.z < m && !vis[tmp.x][tmp.y][tmp.z] && s[tmp.x][tmp.y][tmp.z] != '#')
            {
                que.push(tmp);
                vis[tmp.x][tmp.y][tmp.z] = 1;
            }
        }
    }
    return -1;
}
int main()
{
    while (cin >> r >> n >> m)
    {
        if (r == 0 && n == 0 && m == 0) break;
        memset(vis, 0, sizeof(vis));
        while (!que.empty())
        {
            que.pop();
        }
        for (int i = 0; i < r; i++) {
            string ret;
            for (int j = 0; j < n; j++) {
                cin >> s[i][j];
                for (int k = 0; k < m; k++) {
                    if (s[i][j][k] == 'S') {
                        Begin.x = i, Begin.y = j, Begin.z = k; Begin.t = 0;
                    }
                    else if (s[i][j][k] == 'E') {
                        End.x = i, End.y = j, End.z = k; End.t = 0;
                    }
                }
            }
        }
        int ans = bfs();
        if (ans == -1) {
            printf("Trapped!
");
        }
        else {
            printf("Escaped in %d minute(s).
", ans);
        }
    }

    return 0;
}
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原文地址:https://www.cnblogs.com/xuxiaojin/p/9406408.html