Maze Problem

题目描述

Given a maze, find a shortest path from start to goal.

输入要求

Input consists serveral test cases.

First line of the input contains number of test case T.

For each test case the first line contains two integers N , M ( 1 <= N, M <= 100 ).

Each of the following N lines contain M characters. Each character means a cell of the map.

Here is the definition for chracter.

Constraint:

  • For a character in the map:

    • ‘S’ : start cell

    • ‘E’ : goal cell

    • ‘-‘ : empty cell

    • ‘#’ : obstacle cell

  • no two start cell exists.

  • no two goal cell exists.

输出要求

For each test case print one line containing shortest path. If there exists no path from start to goal, print -1.

假如输入

1
5 5
S-###
-----
##---
E#---
---##

应当输出

9

分析:迷宫最短路径问题,安安分分用bfs宽搜就好。

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#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<vector>
#include<cctype>
#include<cmath>
#include<cstring>
#include<queue>
#include<string>
using namespace std;
struct point
{
int x,y;
}r[10005];
int dis[4][2]={{1,0},{0,-1},{-1,0},{0,1}};
char map[105][105];
int ans[105][105];
int m,n,si,sj,ei,ej;
void bfs()
{
int tail=1,head=0,i,x1,y1;
r[0].x=si;
r[0].y=sj;
while(tail != head)
{
x1=r[head].x;
y1=r[head].y;
for(i=0; i<4; i++)
{
x1+=dis[i][0], y1+=dis[i][1];
if(x1>=0&&y1>=0&&x1<n&&y1<m&&map[x1][y1]!='#'&&ans[x1][y1]==-1)
{
r[tail].x=x1;
r[tail].y=y1;
ans[x1][y1] = 1 + ans[x1-dis[i][0]][y1-dis[i][1]];
tail++;
}
x1-=dis[i][0], y1-=dis[i][1];
}
head++;
if(ans[ei][ej]!=-1)
break;
}
}
void solve();
int main()
{
solve();
return 0;
}
void solve()
{
int t,i,j;
cin>>t;
while(t--)
{
cin>>n>>m;
memset(map,'#',sizeof(map));
memset(ans,-1,sizeof(ans));
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
cin>>map[i][j];
if(map[i][j]=='S')
si=i,sj=j,ans[i][j]=0;
if(map[i][j]=='E')
ei=i,ej=j;
}
getchar();
}
bfs();
if(ans[ei][ej]==-1)
cout<<-1<<endl;
else
cout<<ans[ei][ej]<<endl;
}
}