UVa 1600 Patrol Robot

最新推荐文章于 2020-08-13 14:26:23 发布

神探特斯拉

最新推荐文章于 2020-08-13 14:26:23 发布

阅读量375

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/u013555159/article/details/51317961

本文探讨了在一个含障碍物的网格中，机器人如何寻找从起点到终点的最短路径，特别关注在有限次穿越障碍的情况下，利用深度优先搜索算法解决问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A robot has to patrol around a rectangular area which is in a form of m x n grid (m rows and n columns). The rows are labeled from 1 to m. The columns are labeled from 1 to n. A cell (i, j) denotes the cell in row i and column j in the grid. At each step, the robot can only move from one cell to an adjacent cell, i.e. from (x, y) to (x + 1, y), (x, y + 1), (x - 1, y) or (x, y - 1). Some of the cells in the grid contain obstacles. In order to move to a cell containing obstacle, the robot has to switch to turbo mode. Therefore, the robot cannot move continuously to more than k cells containing obstacles.

Your task is to write a program to find the shortest path (with the minimum number of cells) from cell (1, 1) to cell (m, n). It is assumed that both these cells do not contain obstacles.

Input

The input consists of several data sets. The first line of the input file contains the number of data sets which is a positive integer and is not bigger than 20. The following lines describe the data sets.

For each data set, the first line contains two positive integer numbers m and n separated by space (1 $\le$ m, n $\le$ 20). The second line contains an integer number k (0 $\le$ k $\le$ 20). The i^th line of the next m lines contains n integer a_ij separated by space (i = 1, 2,..., m;j = 1, 2,..., n). The value of a_ij is 1 if there is an obstacle on the cell (i, j), and is 0 otherwise.

Output

For each data set, if there exists a way for the robot to reach the cell (m, n), write in one line the integer number s, which is the number of moves the robot has to make; -1 otherwise.

Sample Input

3 
2 5 
0 
0 1 0 0 0 
0 0 0 1 0 
4 6 
1 
0 1 1 0 0 0
0 0 1 0 1 1
0 1 1 1 1 0
0 1 1 1 0 0
2 2 
0 
0 1 
1 0

Sample Output

7 
10 
-1

一.分析

本问题是在一个权值平均的图上搜索最短路径，应该用bfs来做比较快速，但是一开始没有想到的一个变量来储存还可以穿过几个墙的方法用bfs没有办法解决，所以使用的dfs来解毫无疑问的又超时了。。。。。。。。

二.代码

#include <iostream>
#include <cstring>
using namespace std;
int m,n,k,map[23][23],flag=1,minnum=9999999;
int dx[]={-1,1,0,0};
int dy[]={0,0,-1,1};
void print ()
{
	for(int i=1;i<=m;i++)
	{
		for(int j=1;j<=n;j++)
		{
			cout<<map[i][j]<<" ";
		}
		cout<<endl;
	}
	cout<<endl;
}
int avil(int x,int y,int tk)
{
	if(map[x][y]==-1||x>m||x<=0||y>n||y<=0||(tk==0&&map[x][y]==1)) return 0;
	else return 1;
}
void dfs(int tk,int x,int y,int counter)
{
	int tempk,tempx,tempy,tempc,tempm;
	//tempm=map[x][y];
	for(int i=0;i<=3;i++)
	{
		tempx=x+dx[i],tempy=y+dy[i];
		if(avil(tempx,tempy,tk)==1)
		{
			tempm=map[tempx][tempy];
			tempc=counter+1;
			if(map[tempx][tempy]==1) tempk=tk-1;
			else tempk=k;
			map[tempx][tempy]=-1;
			if(tempx==m&&tempy==n)
			{
				if(counter<=minnum) minnum=tempc;
//				if(tempc==8){
//				cout<<"this is"<<tempc<<endl;
//				print();}
				flag=0;
			}
			else dfs(tempk,tempx,tempy,tempc);
			map[tempx][tempy]=tempm;
		}
	}
}
int main()
{
//	freopen("input.txt","r",stdin);
	int T;
	cin>>T;
	for(int t=1;t<=T;t++)
	{
		cin>>m>>n;
		cin>>k;
		flag=1;
		minnum=999999;
		for(int i=1;i<=m;i++)
		{
			for(int j=1;j<=n;j++)
			{
				cin>>map[i][j];
			}
		}
		//print();
		map[1][1]=-1;
		dfs(k,1,1,0);
		if(flag==1) cout<<"-1"<<endl;
		else cout<<minnum<<endl;
	}
	return 0;
}