Children of the Candy Corn
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10739 | Accepted: 4626 |
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
题目叫你求三个距离：1、总是扶墙往左走。2、总是扶墙往右走。3、最短的距离。
前两个其实就是一样的，可以用dfs。求最短就可以用bfs。
总之是一题让我改了好久的题目。。。。。。
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
char mat[45][45];
int vis[45][45],si,sj,ei,ej,w,h,dir[4][2]={1,0,0,-1,0,1,-1,0};
int flag,rcnt,lcnt;
int fangxiang[]={1,3,0,2};
struct node{
int x,y,step;
};
int check(int x,int y)
{
if(x>=1&&y>=1&&x<=h&&y<=w&&mat[x][y]!=‘#‘&&!vis[x][y])
return 1;
return 0;
}
int checker(int x,int y)
{
if(x>=1&&y>=1&&x<=h&&y<=w)return 1;
return 0;
}
int bfs(int x,int y)
{
node p,q;
queue<node>Q;
p.x=x;
p.y=y;
p.step=1;
Q.push(p);
while(!Q.empty())
{
p=Q.front();
Q.pop();
if(p.x==ei&&p.y==ej)return p.step;
for(int i=0;i<4;i++)
{
q=p;
q.x=p.x+dir[i][0];
q.y=p.y+dir[i][1];
if(check(q.x,q.y)){
q.step++;
vis[q.x][q.y]=1;
Q.push(q);
}
}
}
return 0;
}
void dfs(int x,int y,int cnt,int step) //扶墙往左走：cnt计步数，step是方向
{
if(flag)return;
// printf("1");
if(x==ei&&y==ej){
flag=1;lcnt=cnt; return;
}
int k,i;
switch(step)
{
case 1://左
if(mat[x+1][y]!=‘#‘&&checker(x+1,y)) dfs(x+1,y,cnt+1,4); //这里有一个判断是否出边界的checker函数
else if(mat[x][y-1]!=‘#‘&&checker(x,y-1)) dfs(x,y-1,cnt+1,1);
else if(mat[x-1][y]!=‘#‘&&checker(x-1,y)) dfs(x-1,y,cnt+1,2);
else if(mat[x][y+1]!=‘#‘&&checker(x,y+1)) dfs(x,y+1,cnt+1,3);
break;
case 2://上
if(mat[x][y-1]!=‘#‘&&checker(x,y-1)) dfs(x,y-1,cnt+1,1);
else if(mat[x-1][y]!=‘#‘&&checker(x-1,y)) dfs(x-1,y,cnt+1,2);
else if(mat[x][y+1]!=‘#‘&&checker(x,y+1)) dfs(x,y+1,cnt+1,3);
else if(mat[x+1][y]!=‘#‘&&checker(x+1,y)) dfs(x+1,y,cnt+1,4);
break;
case 3://右
if(mat[x-1][y]!=‘#‘&&checker(x-1,y)) dfs(x-1,y,cnt+1,2);
else if(mat[x][y+1]!=‘#‘&&checker(x,y+1)) dfs(x,y+1,cnt+1,3);
else if(mat[x+1][y]!=‘#‘&&checker(x+1,y)) dfs(x+1,y,cnt+1,4);
else if(mat[x][y-1]!=‘#‘&&checker(x,y-1)) dfs(x,y-1,cnt+1,1);
case 4://下
if(mat[x][y+1]!=‘#‘&&checker(x,y+1)) dfs(x,y+1,cnt+1,3);
else if(mat[x+1][y]!=‘#‘&&checker(x+1,y)) dfs(x+1,y,cnt+1,4);
else if(mat[x][y-1]!=‘#‘&&checker(x,y-1)) dfs(x,y-1,cnt+1,1);
else if(mat[x-1][y]!=‘#‘&&checker(x-1,y)) dfs(x-1,y,cnt+1,2);
break;
}
}
void dfs2(int x,int y,int cnt,int step) //扶墙往右走
{
if(flag)return;
if(x==ei&&y==ej){
flag=1;rcnt=cnt;return;
}
switch(step)
{
case 1://right
if(mat[x+1][y]!=‘#‘&&checker(x+1,y))dfs2(x+1,y,cnt+1,4);
else if(mat[x][y+1]!=‘#‘&&checker(x,y+1))dfs2(x,y+1,cnt+1,1);
else if(mat[x-1][y]!=‘#‘&&checker(x-1,y))dfs2(x-1,y,cnt+1,2);
else if(mat[x][y-1]!=‘#‘&&checker(x,y-1))dfs2(x,y-1,cnt+1,3);
break;
case 2://up
if(mat[x][y+1]!=‘#‘&&checker(x,y+1))dfs2(x,y+1,cnt+1,1);
else if(mat[x-1][y]!=‘#‘&&checker(x-1,y))dfs2(x-1,y,cnt+1,2);
else if(mat[x][y-1]!=‘#‘&&checker(x,y-1))dfs2(x,y-1,cnt+1,3);
else if(mat[x+1][y]!=‘#‘&&checker(x+1,y))dfs2(x+1,y,cnt+1,4);
break;
case 3://left
if(mat[x-1][y]!=‘#‘&&checker(x-1,y))dfs2(x-1,y,cnt+1,2);
else if(mat[x][y-1]!=‘#‘&&checker(x,y-1))dfs2(x,y-1,cnt+1,3);
else if(mat[x+1][y]!=‘#‘&&checker(x+1,y))dfs2(x+1,y,cnt+1,4);
else if(mat[x][y+1]!=‘#‘&&checker(x,y+1))dfs2(x,y+1,cnt+1,1);
case 4://down
if(mat[x][y-1]!=‘#‘&&checker(x,y-1))dfs2(x,y-1,cnt+1,3);
else if(mat[x+1][y]!=‘#‘&&checker(x+1,y))dfs2(x+1,y,cnt+1,4);
else if(mat[x][y+1]!=‘#‘&&checker(x,y+1))dfs2(x,y+1,cnt+1,1);
else if(mat[x-1][y]!=‘#‘&&checker(x-1,y))dfs2(x-1,y,cnt+1,2);
break;
}
}
int main()
{
int n,i,j,turn;
scanf("%d",&n);
while(n--)
{
rcnt=0;lcnt=0;
memset(vis,0,sizeof(vis));
flag=0;
scanf("%d%d",&w,&h);
for(i=1;i<=h;i++)
for(j=1;j<=w;j++)
{
cin>>mat[i][j];
if(mat[i][j]==‘S‘){
si=i;sj=j;
}
if(mat[i][j]==‘E‘){
ei=i;ej=j;
}
}
int ans;
ans=bfs(si,sj);
memset(vis,0,sizeof(vis));
flag=0;
dfs(si,sj,1,1);
flag=0;
dfs2(si,sj,1,1);
printf("%d %d ",lcnt,rcnt);
printf("%d\n",ans);
}
}