poj3083——dfs+bfs综合题

POJ 3083   dfs+bfs+模拟

Children of the Candy Corn

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10564		Accepted: 4539

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

题意：从迷宫起点S到终点E，分别求沿左边的墙走的步数，沿右边墙走的步数，最短路的步数；
思路：最短路直接bfs，沿着墙走dfs+模拟，每次递归时记录方向和并更改计算下一时刻的方向，按方向顺时针（或逆时针）走，注意取余；
难点：模拟。。。弄了3个多小时。。。
bfs还忘了改vis判重造成MLE。。。。模拟时还把左右方向弄反了。。。
唯一欣慰的是0秒过了。。。

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cstdlib>

#include<cstring>

#include<algorithm>

#include<map>

#include<vector>

#include<queue>



using namespace std;



const int maxn=43;



int n,m;

char ch[maxn][maxn];

bool vis[maxn][maxn];

int si,sj;

int ei,ej;

int stepL,stepR,stepS;

struct node

{

    int x,y;

};

int dx[]={0,1,0,-1};

int dy[]={-1,0,1,0};

bool flag;







void bfs(int si,int sj)

{

    queue<node> q;

    q.push({si,sj});

    int dist[maxn][maxn];

    memset(dist,0,sizeof(dist));

    dist[si][sj]=1;

    vis[si][sj]=1;

    while(!q.empty()){

        node now=q.front();

        q.pop();

        for(int i=0;i<4;i++){

            int nx=now.x+dx[i],ny=now.y+dy[i];

            if(vis[nx][ny]||ch[nx][ny]=='#') continue;

            vis[nx][ny]=1;

            dist[nx][ny]=dist[now.x][now.y]+1;

            if(ch[nx][ny]=='E'){

                stepS=dist[nx][ny];

                return;

            }

            q.push({nx,ny});

        }

    }

}



void dfs(int x,int y,char end,int dir,int step,bool tag)

{

    if(flag) return;

    if(ch[x][y]==end){

        if(tag) stepL=step;

        else stepR=step;

        flag=1;return;

    }

    for(int i=0;i<4;i++){

        int nx=x+dx[(dir+i)%4],ny=y+dy[(dir+i)%4];

        int ndir=(dir+i+3)%4;

        if(ch[nx][ny]=='#') continue;

        dfs(nx,ny,end,ndir,step+1,tag);

    }

}



int main()

{

    int T;cin>>T;

    while(T--){

        cin>>m>>n;

        stepL=stepR=stepS=0;

        memset(ch,'#',sizeof(ch));

        for(int i=1;i<=n;i++){

            for(int j=1;j<=m;j++){

                cin>>ch[i][j];

                if(ch[i][j]=='S'){

                    si=i;sj=j;

                }

                if(ch[i][j]=='E'){

                    ei=i;ej=j;

                }

            }

        }

        flag=0;

        dfs(si,sj,'E',0,1,1);

        flag=0;

        dfs(ei,ej,'S',0,1,0);

        memset(vis,0,sizeof(vis));

        bfs(si,sj);

        cout<<stepR<<" "<<stepL<<" "<<stepS<<endl;

    }

    return 0;

}

poj3083