Children of the Candy Corn （dfs+bfs）

Children of the Candy Corn

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 72   Accepted Submission(s) : 27

Problem 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'. <br><br>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 ('#'). <br><br>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

 

Source

PKU

 

题意：

有一个迷宫，题目给出了起始位置，让我们分别求出始终贴着左墙和右墙以及从起点到终点最小的步数。

思路：

最少步数好求，直接bfs即可。难就难在两个深搜上面。我们在深搜的时候用face记录现在在朝哪个方向，对于不同方向，按不同顺序走，然后再根据此方向来确定下一步，是继续走还是转弯！

代码：

#include <iostream>#include <stdio.h>#include <cmath>#include <algorithm>#include <cstring>#include <queue>using namespace std;int w,h,flag;int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};bool vis[101][101];struct node{int x,y;int moves;};int ansl,ansr;int mp[101][101];int sx,sy,ex,ey;void dfsl(int x,int y,int step,int face)   //沿着左，1是往左，2是往上，3是往右，4是往下{    if(flag==1)return;    if(x==ex&&y==ey){flag=1;ansl=step;return;}    if(face == 1){                             //往左，则下面是墙，优先搜索 下，左，上，右        if(mp[x+1][y])    dfsl(x+1,y,step+1,4);        if(mp[x][y-1])    dfsl(x,y-1,step+1,1);        if(mp[x-1][y])    dfsl(x-1,y,step+1,2);        if(mp[x][y+1])    dfsl(x,y+1,step+1,3);    }    else if(face == 2){                      //往上，则左面是墙，优先搜索 左 ，上，右，下        if(mp[x][y-1])    dfsl(x,y-1,step+1,1);        if(mp[x-1][y])    dfsl(x-1,y,step+1,2);        if(mp[x][y+1])    dfsl(x,y+1,step+1,3);        if(mp[x+1][y])    dfsl(x+1,y,step+1,4);    }    else if(face == 3){                      //往右，则上面是墙，优先搜索 上，右，下，左        if(mp[x-1][y])    dfsl(x-1,y,step+1,2);        if(mp[x][y+1])    dfsl(x,y+1,step+1,3);        if(mp[x+1][y])    dfsl(x+1,y,step+1,4);        if(mp[x][y-1])    dfsl(x,y-1,step+1,1);    }    else{        if(mp[x][y+1])    dfsl(x,y+1,step+1,3);  //下，则右面是墙，优先搜索 右，下，左，上        if(mp[x+1][y])    dfsl(x+1,y,step+1,4);        if(mp[x][y-1])    dfsl(x,y-1,step+1,1);        if(mp[x-1][y])    dfsl(x-1,y,step+1,2);    }}void dfsr(int x,int y,int step,int face) //沿着右，
1是往左，2是往上，3是往右，4是往下
{ if(flag==1)return; if(x==ex&&y==ey){flag=1;ansr=step;return;} if(face == 1){ //沿着左，则上面是墙，优先搜索 上 ，左，下，右 if(mp[x-1][y]) dfsr(x-1,y,step+1,2); if(mp[x][y-1]) dfsr(x,y-1,step+1,1); if(mp[x+1][y]) dfsr(x+1,y,step+1,4); if(mp[x][y+1]) dfsr(x,y+1,step+1,3); } else if(face == 2){ //沿着上，则右边是墙，优先搜索 右，上，左，下 if(mp[x][y+1]) dfsr(x,y+1,step+1,3); if(mp[x-1][y]) dfsr(x-1,y,step+1,2); if(mp[x][y-1]) dfsr(x,y-1,step+1,1); if(mp[x+1][y]) dfsr(x+1,y,step+1,4); } else if(face == 3){ //沿着右，则下面是墙，优先搜索 下，右，上，左 if(mp[x+1][y]) dfsr(x+1,y,step+1,4); if(mp[x][y+1]) dfsr(x,y+1,step+1,3); if(mp[x-1][y]) dfsr(x-1,y,step+1,2); if(mp[x][y-1]) dfsr(x,y-1,step+1,1); } else{ //沿着下走，则左面是墙，优先搜索 左，下，右，上 if(mp[x][y-1]) dfsr(x,y-1,step+1,1); if(mp[x+1][y]) dfsr(x+1,y,step+1,4); if(mp[x][y+1]) dfsr(x,y+1,step+1,3); if(mp[x-1][y]) dfsr(x-1,y,step+1,2); }}int bfs() //求最短路径{ queue<node>q; vis[sx][sy]=1; node t,tt,start; start.x=sx; start.y=sy; start.moves=1; q.push(start); int sum; while(!q.empty()) { t=q.front(); q.pop(); if(t.x== ex && t.y==ey){ sum=t.moves; } for(int i=0;i<4;i++) { int xxx=t.x+dir[i][0]; int yyy=t.y+dir[i][1]; if(mp[xxx][yyy]==1&&!vis[xxx][yyy]&&xxx>=1&&xxx<=h&&yyy>=1&&yyy<=w){vis[xxx][yyy]=1;tt.x=xxx,tt.y=yyy,tt.moves=t.moves+1;q.push(tt);} } } return sum;}int main(){ //freopen("D:/a.txt","r",stdin); int t; int i,j; char str[50]; cin>>t; while(t--) { memset(mp,0,sizeof(mp)); cin>>w>>h; getchar(); for(i=1;i<=h;i++) { gets(str); for(j=0;j<w;j++) { if(str[j]=='S'){mp[i][j+1]=1;sx=i,sy=j+1;} if(str[j]=='E'){mp[i][j+1]=1;ex=i,ey=j+1;} if(str[j]=='.'){mp[i][j+1]=1;} } } memset(vis,0,sizeof(vis)); flag=0;dfsl(sx,sy,1,1); flag=0;dfsr(sx,sy,1,1); cout<<ansl<<' '<<ansr<<' '; cout<<bfs()<<endl; } return 0;}
心得：
花费了好几个小时。。。能力不足啊