Children of the Candy Corn
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 11215 | Accepted: 4841 |
Description
The cornfield maze is a popular Halloween treat. Visitors are shown the entrance and must wander through the maze facing zombies, chainsaw-wielding psychopaths, hippies, and other terrors on their quest to find the exit.
One popular maze-walking strategy guarantees that the visitor will
eventually find the exit. Simply choose either the right or left wall,
and follow it. Of course, there‘s no guarantee which strategy (left or
right) will be better, and the path taken is seldom the most efficient.
(It also doesn‘t work on mazes with exits that are not on the edge;
those types of mazes are not represented in this problem.)
As the proprieter of a cornfield that is about to be converted into a
maze, you‘d like to have a computer program that can determine the left
and right-hand paths along with the shortest path so that you can
figure out which layout has the best chance of confounding visitors.
Input
Input
to this problem will begin with a line containing a single integer n
indicating the number of mazes. Each maze will consist of one line with a
width, w, and height, h (3 <= w, h <= 40), followed by h lines of
w characters each that represent the maze layout. Walls are represented
by hash marks (‘#‘), empty space by periods (‘.‘), the start by an ‘S‘
and the exit by an ‘E‘.
Exactly one ‘S‘ and one ‘E‘ will be present in the maze, and they
will always be located along one of the maze edges and never in a
corner. The maze will be fully enclosed by walls (‘#‘), with the only
openings being the ‘S‘ and ‘E‘. The ‘S‘ and ‘E‘ will also be separated
by at least one wall (‘#‘).
You may assume that the maze exit is always reachable from the start point.
Output
For
each maze in the input, output on a single line the number of (not
necessarily unique) squares that a person would visit (including the ‘S‘
and ‘E‘) for (in order) the left, right, and shortest paths, separated
by a single space each. Movement from one square to another is only
allowed in the horizontal or vertical direction; movement along the
diagonals is not allowed.
Sample Input
2 8 8 ######## #......# #.####.# #.####.# #.####.# #.####.# #...#..# #S#E#### 9 5 ######### #.#.#.#.# S.......E #.#.#.#.# #########
Sample Output
37 5 5 17 17 9
Source
题目大意：从入口S进到出口E，如果一直按照走左边的路（碰到墙后返回来继续），或一直选择向右走，各自需要多少步；和求最短路（简单BFS）
这道题虽说是一道题，却同时考察了DFS和BFS
当然，这道题的困难之处不在于DFS有多复杂，而在于如何一直向左走或向右走却不受位置的影响，解决办法是设置两个数组顺时针与逆时针，方法如下：
设左上右下为 0， 1， 2， 3 顺时针时，假设当前的前进方向为d， 那么从(d+2)%4，也就是相反方向开始循环，每次 (d+1)%4，遇到第一个能走的就前进。 逆时针时同理，不同的是每次(d-1+4)%4。下面附上本人比较挫代码
#include<stdio.h>
#include<string.h>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=100;
char str[maxn][maxn];
bool vis[maxn][maxn];
int stepl=0,stepr=0,stepm=0;
int n,m;
int sx,sy,sd;
int dirl[4][2]={-1,0,0,1,1,0,0,-1};
int dirr[4][2]={1,0,0,1,-1,0,0,-1};
struct node{
int x,y,dis;
};
node u,v;
int step;
int dfs(int x,int y,int dir,int a[][2]){
if(str[x][y]==‘E‘)
return 1;//特别注意，返回1
for(int i=0;i<4;i++){
int tdir=(dir+i+3)%4;//tdir需要重新定义
int tx=x+a[tdir][0];
int ty=y+a[tdir][1];
if(tx>=0&&tx<n&&ty>=0&&ty<m&&str[tx][ty]!=‘#‘&&!vis[tx][ty]){
step =dfs(tx,ty,tdir,a)+1;//+1
break;
}
}
return step;
}
int bfs(){
memset(vis,false,sizeof(vis));
u.x=sx,u.y=sy,u.dis=1;
vis[sx][sy]=true;
queue<node>q;
q.push(u);
while(!q.empty()){
u=q.front();
q.pop();
if(str[u.x][u.y]==‘E‘)
return u.dis;
for(int i=0;i<4;i++){
v.x=u.x+dirl[i][0];
v.y=u.y+dirl[i][1];
if(v.x>=0&&v.x<n&&v.y>=0&&v.y<m&&
!vis[v.x][v.y]&&str[v.x][v.y]!=‘#‘){
vis[v.x][v.y]=true;
v.dis=u.dis+1;
q.push(v);
}
}
}
}
int main(){
int t;
scanf("%d",&t);
while(t--){
memset(str,0,sizeof(str));
memset(vis,false,sizeof(vis));
scanf("%d%d",&m,&n);
for(int i=0;i<n;i++){
scanf("%s",str[i]);
for(int j=0;j<m;j++){
if(str[i][j]==‘S‘)
sx=i,sy=j;
}
}
int tx,ty;
for(int i=0;i<4;i++){//判断左走方向
tx=sx+dirl[i][0];
ty=sy+dirl[i][1];
if(str[tx][ty]==‘.‘){
sd=i;
break;
}
}
step=0;//每次向左或向右需要重新赋值为9
stepl=dfs(sx,sy,sd,dirl);//对于左走进行dfs
for(int i=0;i<4;i++){//判断左走方向
tx=sx+dirr[i][0];
ty=sy+dirr[i][1];
if(str[tx][ty]==‘.‘){
sd=i;
break;
}
}
step=0;
stepr=dfs(sx,sy,sd,dirr);//对于左走进行dfs
stepm=bfs();//最短路进行bfs
printf("%d %d %d\n",stepl,stepr,stepm);
}
return 0;
}