POJ 2688 Cleaning Robot (BFS+DFS)

最新推荐文章于 2021-12-22 22:04:04 发布

转载最新推荐文章于 2021-12-22 22:04:04 发布 · 183 阅读

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/joeylee97/p/7017577.html

本文探讨了一种针对移动机器人在充满家具的矩形房间中清洁地面的路径规划算法。算法的目标是最小化从初始位置到达所有脏瓷砖所需的移动次数，同时考虑到房间中可能存在的障碍物。通过广度优先搜索和图的遍历策略，该算法确定了机器人能否及如何访问所有脏瓷砖，以实现最高效的清洁任务。

Cleaning Robot

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4952		Accepted: 1921

Description

Here, we want to solve path planning for a mobile robot cleaning a rectangular room floor with furniture.

Consider the room floor paved with square tiles whose size fits the cleaning robot (1 * 1). There are 'clean tiles' and 'dirty tiles', and the robot can change a 'dirty tile' to a 'clean tile' by visiting the tile. Also there may be some obstacles (furniture) whose size fits a tile in the room. If there is an obstacle on a tile, the robot cannot visit it. The robot moves to an adjacent tile with one move. The tile onto which the robot moves must be one of four tiles (i.e., east, west, north or south) adjacent to the tile where the robot is present. The robot may visit a tile twice or more.

Your task is to write a program which computes the minimum number of moves for the robot to change all 'dirty tiles' to 'clean tiles', if ever possible.

Input

The input consists of multiple maps, each representing the size and arrangement of the room. A map is given in the following format.

w h
c11 c12 c13 ... c1w
c21 c22 c23 ... c2w
...
ch1 ch2 ch3 ... chw

The integers w and h are the lengths of the two sides of the floor of the room in terms of widths of floor tiles. w and h are less than or equal to 20. The character cyx represents what is initially on the tile with coordinates (x, y) as follows.

'.' : a clean tile
'*' : a dirty tile
'x' : a piece of furniture (obstacle)
'o' : the robot (initial position)

In the map the number of 'dirty tiles' does not exceed 10. There is only one 'robot'.

The end of the input is indicated by a line containing two zeros.

Output

For each map, your program should output a line containing the minimum number of moves. If the map includes 'dirty tiles' which the robot cannot reach, your program should output -1.

Sample Input

7 5
.......
.o...*.
.......
.*...*.
.......
15 13
.......x.......
...o...x....*..
.......x.......
.......x.......
.......x.......
...............
xxxxx.....xxxxx
...............
.......x.......
.......x.......
.......x.......
..*....x....*..
.......x.......
10 10
..........
..o.......
..........
..........
..........
.....xxxxx
.....x....
.....x.*..
.....x....
.....x....
0 0

Sample Output

8
49
-1

Source

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN 21
#define N 33
#define MOD 1000000
#define INF 1000000009
const double eps = 1e-9;
const double PI = acos(-1.0);
/*
给定障碍，起始点 
问最少花费多少时间遍历所有点
*/
int m, n, index[MAXN][MAXN], ans, cnt;
char g[MAXN][MAXN];
bool been[MAXN][MAXN],vis[MAXN];
int X[4] = { -1,0,1,0 }, Y[4] = { 0,1,0,-1 };
struct node
{
    int x, y, t;
    node(int _x,int _y,int _t):x(_x),y(_y),t(_t){}
};
vector<node> pos;
struct edge
{
    int to, cost;
    edge(int _t,int _c):to(_t),cost(_c){}
};
vector<edge> E[MAXN];
bool bfs(int sx, int sy)
{
    memset(been, false, sizeof(been));
    queue<node> q;
    q.push(node(sx, sy, 0));
    been[sx][sy] = true;
    while (!q.empty())
    {
        node t = q.front();
        q.pop();
        if (g[t.x][t.y] == 'o' || g[t.x][t.y] == '*')
        {
            E[index[sx][sy]].push_back(edge(index[t.x][t.y], t.t));
        }
        for (int i = 0; i < 4; i++)
        {
            int nx = t.x + X[i], ny = t.y + Y[i];
            if (nx >= 0 && ny >= 0 && nx < n&&ny < m && !been[nx][ny] && g[nx][ny] != 'x')
            {
                been[nx][ny] = true;
                q.push(node(nx, ny, t.t + 1));
            }
        }
    }
    if (E[index[sx][sy]].size() != cnt)
        return false;
    else
        return true;
}
void dfs(int k, int time,int tmp)
{
    if (time > ans) return;
    if (tmp == cnt)
    {
        ans = min(time, ans);
        return;
    }
    for (int i = 0; i < E[k].size(); i++)
    {
        if (!vis[E[k][i].to])
        {
            vis[E[k][i].to] = true;
            dfs(E[k][i].to, time + E[k][i].cost,tmp+1);
            vis[E[k][i].to] = false;
        }
    }
}
int main()
{
    while (scanf("%d%d", &m, &n), n + m)
    {
        ans = INF;
        pos.clear();
        for (int i = 0; i < MAXN; i++)
            E[i].clear();
        memset(vis, false, sizeof(vis));
        cnt = 1;
        for (int i = 0; i < n; i++)
        {
            scanf("%s", g[i]);
            for (int j = 0; j < m; j++)
            {
                if ( g[i][j] == '*')
                {
                    index[i][j] = cnt++;
                    pos.push_back(node(i, j, 0));
                }
                else if (g[i][j] == 'o')
                {
                    index[i][j] = 0;
                    pos.insert(pos.begin(),node(i,j,0));
                }
            }
        }
        bool f = false;
        for (int i = 0; i < pos.size(); i++)
        {
            if (!bfs(pos[i].x, pos[i].y))
            {
                f = true;
                break;
            }
        }
        if (f)
        {
            printf("-1\n");
            continue;
        }
        vis[0] = true;
        dfs(0, 0, 1);
        printf("%d\n", ans);
    }
}

转载于:https://www.cnblogs.com/joeylee97/p/7017577.html