Children of the Candy Corn
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 8046   Accepted: 3518

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
#include<iostream>
#include<stdio.h>
using namespace std;
int w,h,ex,ey,sx,sy;
int map[100][100],can[100][100];
struct vid{ int x,y,step;
}queue[5000];
int zan[4][2]={{-1,0},{0,1},{1,0},{0,-1}};
int dirl[4][2]={0,-1,-1,0,0,1,1,0},dirr[4][2]={0,1,1,0,0,-1,-1,0};
int dfsr(int dstep,int x,int y,int di )
{
int i,temp1,temp2; for(i=0;i<4;i++)
{
temp1=x+dirr[i][0];
temp2=y+dirr[i][1]; if((temp1>=0)&&(temp1<h)&&(temp2>=0)&&(temp2<w))
{
if(dfsr(dstep+1,x+dirr[i][0],y+dirr[i][1],i))
return dstep;
else
return 0;
} }
return 0; }
int dfsl(int dstep,int x,int y,int di )
{ int i,temp1,temp2; for(i=0;i<4;i++)
{
temp1=x+dirl[i][0];
temp2=y+dirl[i][1]; if((temp1>=0)&&(temp1<h)&&(temp2>=0)&&(temp2<w))
{
if(dfsl(dstep+1,temp1,temp2,i))
return dstep;
else
return 0;
} } return 0;
}
int bfs()
{ if (sx == ex && sy == ey)
{ return 1;
}
int t,ww,x,y,t1,t2;
t=ww=1;
queue[t].x=sx;
queue[t].y=sy;
queue[t].step=0;
can[sx][sy]=1;
while(t<=ww&&!can[ex][ey])
{ x=queue[t].x;
y=queue[t].y;
for(int i=0;i<4;i++)
if((!map[x+zan[i][0]][y+zan[i][1]])&&(!can[x+zan[i][0]][y+zan[i][1]]))
{ t1=x+zan[i][0];
t2=y+zan[i][1];
if(t1>=0&&(t1<h)&&(t2>=0)&&(t2<w)&&(!map[t1][t2])&&(!can[t1][t2]))
{ queue[++ww].x=t1;
queue[ww].y=t2;
queue[ww].step=queue[t].step+1;
can[t1][t2]=1;
}
}
t++;
}
return queue[ww].step+1;
}
int main ()
{
int t;
char c;
scanf("%d",&t);
getchar();
while(t--)
{ scanf("%d%d",&w,&h);
getchar();
for(int i=0;i<h;i++)
{
for(int j=0;j<w;j++)
{
can[i][j]=0;
c=getchar();
if(c=='#')
map[i][j]=1;
else if(c=='.')
map[i][j]=0;
else if(c=='S')
{
map[i][j]=0;
sx=i,sy=j;
}
else
if(c=='E')
{
map[i][j]=0;
ex=i;ey=j;
}
}
getchar();
}
// init();
// dfsr();
// printf("%d ",bfs());
printf("%d %d %d\n",dfsl(0,sx,sy,0)+1,dfsr(0,sx,sy,0)+1,bfs());
} return 0;
}
												

poj3083的更多相关文章

  1. ACM/ICPC 之 靠墙走-DFS+BFS(POJ3083)

    //POJ3083 //DFS求靠左墙(右墙)走的路径长+BFS求最短路 //Time:0Ms Memory:716K #include<iostream> #include<cst ...

  2. POJ3083——Children of the Candy Corn(DFS+BFS)

    Children of the Candy Corn DescriptionThe cornfield maze is a popular Halloween treat. Visitors are ...

  3. POJ3083 Children of the Candy Corn(搜索)

    题目链接. 题意: 先沿着左边的墙从 S 一直走,求到达 E 的步数. 再沿着右边的墙从 S 一直走,求到达 E 的步数. 最后求最短路. 分析: 最短路好办,关键是沿着墙走不太好想. 但只要弄懂如何 ...

  4. poj3083走玉米地问题

    走玉米地迷宫,一般有两种简单策略,遇到岔路总是优先沿着自己的左手方向,或者右手方向走.给一个迷宫,给出这两种策略的步数,再给出最短路径的长度. ######### #.#.#.#.# S....... ...

  5. POJ3083 Children of the Candy Corn(Bfs + Dfs)

    题意:给一个w*h的迷宫,其中矩阵里面 S是起点,E是终点,“#”不可走,“.”可走,而且,S.E都只会在边界并且,不会在角落,例如(0,0),输出的话,每组数据就输出三个整数,第一个整数,指的是,以 ...

  6. POJ-3083 Children of the Candy Corn (BFS+DFS)

    Description The cornfield maze is a popular Halloween treat. Visitors are shown the entrance and mus ...

  7. DFS+BFS(POJ3083)

    题目链接:http://poj.org/problem?id=3083 解题报告:这个题目,搜最短路,没有什么问题.优先走左边,走右边,有很多说法,思路大概都相同,都是记录当前朝向,根据数学公式(i+ ...

  8. poj3083 Children of the Candy Corn BFS&&DFS

    Children of the Candy Corn Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11215   Acce ...

  9. 基础BFS+DFS poj3083

    //满基础的一道题 //最短路径肯定是BFS. //然后靠右,靠左,就DFS啦 //根据前一个状态推出下一个状态,举靠左的例子,如果一开始是上的话,那么他的接下来依次就是 左,上 , 右 , 下 // ...

随机推荐

  1. sdut 2351 In Danger (找规律)

    题目:http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2351 题意:xyez, xy表示一个十进 ...

  2. UVa 11389 (贪心) The Bus Driver Problem

    题意: 有司机,下午路线,晚上路线各n个.给每个司机恰好分配一个下午路线和晚上路线. 给出行驶每条路线的时间,如果司机开车时间超过d,则要付加班费d×r. 问如何分配路线才能使加班费最少. 分析: 感 ...

  3. HttpContext.Current.RewritePath方法重写URL

    if (!IsPostBack) { //如果请求ID为空,则重写URL为:~/index.aspx?ID=shouji.115sou.com if (Request.QueryString[&quo ...

  4. 【 D3.js 视频系列 】 飞速入门

    本教程共包含 6 个视频,目的是为了帮助初学者快速入门,以便阅读本站其他文章. 本教程的名称为"飞速入门",是为初学者准备的,其中包括了 D3 开发中最基础的知识.对 D3 掌握得 ...

  5. 2013.11.15 初学ant构建

    该做的事情都差不多做完了,今天开始用ant构建,所以学了下ant,其实要不是因为ubuntu时不时的抽风我应该早就可以开始构建了,但重写的时候也想清楚了一些逻辑,优化了一些地方.下面是我这辈子写的第一 ...

  6. 一天一个Java基础——排序

    插入排序 直接插入排序: 当插入第i个数据元素k时,由前i-1个数据元素组成已排序的数据序列,将k与数据序列中各数据元素依次进行比较后,插入到数据序列的适当位置,使得插入后的数据序列仍是排序的. 直接 ...

  7. Android-使用getIdentifier()获取资源Id

    使用getIdentifier()获取资源Id int i= getResources().getIdentifier("icon", "drawable", ...

  8. 关于WCF中间层服务器端DTO属性更新如何同步回仓储实体的处理方式

    中间层建立上下文录制对象及录制属性.如下范例: public bool CancelChangeEvent(ClientContext context, Dbs dbs, int encounterI ...

  9. 2014年acm亚洲区域赛·鞍山站

    今天北京赛站的比赛也结束了···看了一天的直播之后意识到鞍山站的比赛都过去了一个多月了···这一个月比较萎靡···整天都在睡觉写报告画工图中度过··· 鞍山比哈尔滨还是暖和很多的···就是山上有奇怪的 ...

  10. Java SE 6 新特性: 对脚本语言的支持

    2006 年底,Sun 公司发布了 Java Standard Edition 6(Java SE 6)的最终正式版,代号 Mustang(野马).跟 Tiger(Java SE 5)相比,Musta ...