poj3083解題報告

Children of the Candy Corn

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4046		Accepted: 1838

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

28 8#########......##.####.##.####.##.####.##.####.##...#..##S#E####9 5##########.#.#.#.#S.......E#.#.#.#.##########

Sample Output

37 5 517 17 9

題目大意:給定一個迷宮，S是起點，E是終點，#是牆不可走，.可以走。先輸出貼左邊牆走到E的步數，再輸出貼

右邊牆的，再輸出最少走多少部能到E。。。

思路：這個是個綜合題，BFS和DFS全用上了，代碼快3KB了，我靠的，少考慮了一個情況，WA了許久，終於A了 我靠的。。。

8 8
########
#....#.#
#....#.#
#....#.#
#....#.#
#....#.#
S......#
#E######

少考慮了S轉個彎就是E的情況。。WA了許久，終於發現這個BUG了。。。。汗。。

#include<iostream>using namespace std;char map[41][41],lock[41][41];int head,rear,m,n,queue[2000];struct mm{int x,y;}mm[4]={-1,0,0,1,1,0,0,-1};void push(int i){queue[rear++%2000]=i;}int pop(){return queue[head++%2000];}int bfs(){int x,y,step,i;while(head!=rear){x=pop();y=pop();step=pop();if(map[x][y]=='E')return step;for(i=0;i<4;i++)if(x+mm[i].x>-1&&x+mm[i].x<m&&y+mm[i].y>-1&&y+mm[i].y<n&&(map[x+mm[i].x][y+mm[i].y]=='.'||map[x+mm[i].x][y+mm[i].y]=='E')&&lock[x+mm[i].x][y+mm[i].y]==0){push(x+mm[i].x);push(y+mm[i].y);push(step+1);lock[x+mm[i].x][y+mm[i].y]=1;}}return 0;}int ldfs(int x,int y,int f){if(map[x][y]=='E')return 1;if((map[x+mm[f].x][y+mm[f].y]=='.'||map[x+mm[f].x][y+mm[f].y]=='E')&&map[x+mm[(f+3)%4].x][y+mm[(f+3)%4].y]=='#')return 1+ldfs(x+mm[f].x,y+mm[f].y,f);elseif(!(map[x+mm[(f+3)%4].x][y+mm[(f+3)%4].y]=='#')){return 1+ldfs(x+mm[(f+3)%4].x,y+mm[(f+3)%4].y,(f+3)%4);}elsereturn ldfs(x,y,(f+1)%4);}int rdfs(int x,int y,int f){if(map[x][y]=='E')return 1;if((map[x+mm[f].x][y+mm[f].y]=='.'||map[x+mm[f].x][y+mm[f].y]=='E')&&map[x+mm[(f+1)%4].x][y+mm[(f+1)%4].y]=='#')return 1+rdfs(x+mm[f].x,y+mm[f].y,f);elseif(!(map[x+mm[(f+1)%4].x][y+mm[(f+1)%4].y]=='#')){return 1+rdfs(x+mm[(f+1)%4].x,y+mm[(f+1)%4].y,(f+1)%4);}elsereturn rdfs(x,y,(f+3)%4);}int main(){int t,i,j,x,y,f,o;cin>>t;while(t--){memset(lock,0,sizeof(lock));head=rear=0;cin>>n>>m;for(i=0;i<m;i++)cin>>map[i];for(i=0;i<m;i++)for(j=0;j<n;j++)if(map[i][j]=='S')goto start;start:for(f=0;f<4;f++)if(i+mm[f].x>-1&&i+mm[f].x<m&&j+mm[f].y>-1&&j+mm[f].y<n&&(map[i+mm[f].x][j+mm[f].y]=='.'||map[i+mm[f].x][j+mm[f].y]=='E')&&map[i+mm[f].x+mm[(f+3)%4].x][j+mm[f].y+mm[(f+3)%4].y]=='#'){o=1+ldfs(i+mm[f].x,j+mm[f].y,f);break;}else if(i+mm[f].x>-1&&i+mm[f].x<m&&j+mm[f].y>-1&&j+mm[f].y<n&&map[i+mm[f].x][j+mm[f].y]=='.'&&(map[i+mm[f].x+mm[(f+3)%4].x][j+mm[f].y+mm[(f+3)%4].y]=='.'||map[i+mm[f].x+mm[(f+3)%4].x][j+mm[f].y+mm[(f+3)%4].y]=='E')){o=2+ldfs(i+mm[f].x+mm[(f+3)%4].x,j+mm[f].y+mm[(f+3)%4].y,(f+3)%4);break;}cout<<o<<' ';for(f=0;f<4;f++)if(i+mm[f].x>-1&&i+mm[f].x<m&&j+mm[f].y>-1&&j+mm[f].y<n&&(map[i+mm[f].x][j+mm[f].y]=='.'||map[i+mm[f].x][j+mm[f].y]=='E')&&map[i+mm[f].x+mm[(f+1)%4].x][j+mm[f].y+mm[(f+1)%4].y]=='#'){o=1+rdfs(i+mm[f].x,j+mm[f].y,f);break;}else if(i+mm[f].x>-1&&i+mm[f].x<m&&j+mm[f].y>-1&&j+mm[f].y<n&&map[i+mm[f].x][j+mm[f].y]=='.'&&(map[i+mm[f].x+mm[(f+1)%4].x][j+mm[f].y+mm[(f+1)%4].y]=='.'||map[i+mm[f].x+mm[(f+1)%4].x][j+mm[f].y+mm[(f+1)%4].y]=='E')){o=2+rdfs(i+mm[f].x+mm[(f+1)%4].x,j+mm[f].y+mm[(f+1)%4].y,(f+1)%4);break;}cout<<o<<' ';lock[i][j]=1;push(i);push(j);push(1);cout<<bfs()<<endl;}return 0;}