Children of the Candy Corn (bfs+dfs)
2013-08-07 10:37
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Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 8120 | Accepted: 3547 |
InputInput 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.
OutputFor 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<stdio.h> #include<string.h> #include<queue> #include<iostream> using namespace std; struct node { int x,y; int step; }; int n,m; char map[50][50]; int vis[50][50]; int dir[4][2] = {{-1,0},{0,-1},{1,0},{0,1}};//上左下右顺时针; queue <node> que; int ans3; int dfs(int sx,int sy, int ex, int ey, int d, int ans) { if(sx == ex && sy == ey) return ans; d = (d+1)%4;//左转; while(map[sx+dir[d][0]][sy+dir[d][1]] == '#') d = (d+3)%4;//当有墙的时候就右转直到没有墙; return dfs(sx+dir[d][0],sy+dir[d][1],ex,ey,d,ans+1); } void bfs(int sx,int sy, int ex, int ey) { while(!que.empty()) que.pop(); memset(vis,0,sizeof(vis)); que.push((struct node){sx,sy,1}); vis[sx][sy] = 1; while(!que.empty()) { struct node u = que.front(); que.pop(); if(u.x == ex && u.y == ey) { ans3 = u.step; return; } if(u.x-1 >= 1 && !vis[u.x-1][u.y] && map[u.x-1][u.y] != '#') { vis[u.x-1][u.y] = 1; que.push((struct node){u.x-1,u.y,u.step+1}); } if(u.x+1 <= n && !vis[u.x+1][u.y] && map[u.x+1][u.y] != '#') { vis[u.x+1][u.y] = 1; que.push((struct node){u.x+1,u.y,u.step+1}); } if(u.y-1 >= 1&& !vis[u.x][u.y-1] && map[u.x][u.y-1] != '#') { vis[u.x][u.y-1] = 1; que.push((struct node){u.x,u.y-1,u.step+1}); } if(u.y+1 <= m && !vis[u.x][u.y+1] && map[u.x][u.y+1] != '#') { vis[u.x][u.y+1] = 1; que.push((struct node){u.x,u.y+1,u.step+1}); } } } int main() { int t; int sx,sy,ex,ey; scanf("%d",&t); while(t--) { memset(map,'#',sizeof(map));//外面加一道墙,从s进去之后只有一条路可走; scanf("%d %d",&m,&n); for(int i = 1; i <= n; i++) { for(int j = 1; j <= m; j++) { cin>>map[i][j]; if(map[i][j] == 'S') { sx = i; sy= j; } if(map[i][j] == 'E') { ex = i; ey = j; } } } printf("%d %d ",dfs(sx,sy,ex,ey,1,1),dfs(ex,ey,sx,sy,1,1)); bfs(sx,sy,ex,ey); printf("%d\n",ans3); } return 0; }View Code
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