Children of the Candy Corn
Time Limit:1000MSMemory Limit:65536K
Total Submissions:10380Accepted:4485
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
注意：1.使用DFS计算左转优先和右转优先的路径，使用BFS计算最短路径2.这里的DFS不需要标记，因为按照方向顺时针（或逆时针）前进时，除非无路可走才会返回，所以不会因为没有标记而出现问题，不过它的前进方向是相对，是相对当前位置进行左上右下（右前左后）探索（这个相对性非常重要），最初的方向由起点S确定，而下一步的方向则由前一步的走向决定以顺时针，左手优先为例，即（相对于当前方向，左转90°后的方向为第一优先，，依次右转90°找可能的通路）如何左转90度？0,1,2,3代表四个方向，显示直接-1是不行的，可能变成负值了，那么可以用(d+3)%4,就可以左转90度到优先位置，当然往右转的过程中，还是会超过0,1,2,3，到了外面，所以在运算时，使用d%4就可以了。
<pre name="code" class="html">#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>#include <set>#include <queue>#include <stack>using namespace std;struct node {int x,y;int step;};char map[50][50];int vis[50][50];int jx[]= {0,1,0,-1};int jy[]= {1,0,-1,0};//设置为右上左下int jr[]= {1,0,3,2};int jl[]= {3,0,1,2};int cnt,d;char sx,sy,ex,ey;int w,h;void dfs_left(int d,int x,int y){int i,j;if(x==ex&&y==ey) {printf("%d ",cnt);return ;}cnt++;for(j=0; j<4; j++) {i=(d+jl[j])%4;int dx=x+jx[i];int dy=y+jy[i];if(dx>=0&&dx<h&&dy>=0&&dy<w&&map[dx][dy]!='#') {dfs_left(i,dx,dy);return;}}}void dfs_right(int d,int x,int y){int i,j;if(x==ex&&y==ey) {printf("%d ",cnt);return ;}cnt++;for(j=0; j<4; j++) {i=(d+jr[j])%4;int dx=x+jx[i];int dy=y+jy[i];if(dx>=0&&dx<h&&dy>=0&&dy<w&&map[dx][dy]!='#') {dfs_right(i,dx,dy);return;}}}void bfs(int x,int y){int i;memset(vis,0,sizeof(vis));queue<node >q;struct node t,f;t.x=x;t.y=y;t.step=0;q.push(t);vis[t.x][t.y]=1;while(!q.empty()) {t=q.front();q.pop();if(t.x==ex&&t.y==ey) {printf("%d\n",t.step+1);return ;}for(i=0; i<4; i++) {f.x=t.x+jx[i];f.y=t.y+jy[i];if(f.x>=0&&f.x<h&&f.y>=0&&f.y<w&&map[f.x][f.y]!='#'&&!vis[f.x][f.y]) {vis[f.x][f.y]=1;f.step=t.step+1;q.push(f);}}}}int main(){int T,i,j;scanf("%d",&T);while(T–) {scanf("%d %d",&w,&h);memset(vis,0,sizeof(vis));for(i=0; i<h; i++) {scanf("%s",map[i]);for(j=0; j<w; j++) {if(map[i][j]=='S') {sx=i;sy=j;}if(map[i][j]=='E') {ex=i;ey=j;}}}if(sx==0) d=0;if(sx==h-1) d=2;if(sy==0) d=1;if(sy==w-1) d=3;cnt=1;dfs_left(d,sx,sy);cnt=1;dfs_right(d,sx,sy);bfs(sx,sy);}return 0;}
风景如何，其实并不重要。重要的是，你在我的身边。
