POJ-3083 Children of the Candy Corn problem-solving report (search) wide search deep search Maze __output

B-children of the Candy CornTime Limit:1000MSMemory Limit:65536KB64bit IO Format:%I64D &%i64u http://openoj.awaysoft.com:8080/judge/problem/viewProblem.action?id=28859 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, a nd other terrors on their quest to find the exit.

One popular maze-walking strategy guarantees that visitor would eventually find the exit. Simply choose either the right or left wall, and follow it. Course, there ' s no guarantee which strategy (left or right) would be better, and the path taken is seldom the most effic Ient. (It also doesn ' t work on mazes with exits "that is are not on the edge; those types of mazes are not represented at this probl EM.)

As the proprieter of a cornfield that's about to being converted into a maze, your ' d like to have a computer program that can Determine the left and right-hand paths along with the shortest path so, can figure out which layout has the best Chance of confounding visitors.

Input input to this problem would begin with a-line containing a single integer n indicating the number of mazes. Each maze would consist of one line with a width, w, and Height, h (3 <= W, H <=), followed by H lines of W Charac Ters each of that represent the maze layout. Walls are represented by hash marks (' # '), Empty spaces by periods ('. '), the "Start by" an ' S ' and "Exit by" an ' E '.

exactly one ' S ' and one ' E ' would be present in the maze, and they'll always be located along one of the maze E Dges and never in a corner. The maze is fully enclosed by walls (' # '), with the only openings being the ' S ' and ' E '. The ' S ' and ' E ' would also be separated by at least one wall (' # ').

You may assume this maze exit is always reachable to 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 wo Uld visit (including the ' S ' and ' E ') for [in order] the left, right, and shortest paths, separated by a single spaces each . Movement from one square to another are 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

Panax 5 5
17 17 9

Hint

Finally the topic meaning to see understand = =.

Left first: Each will be a step, priority to walk it in the left direction; Right first: Every step of the line, priority to walk its right direction;

If the priority direction does not work, then according to the diagram direction in turn to judge;

Source of report: http://blog.csdn.net/angle555945/article/details/7303207

#include <stdio.h> #include <memory.h> const int maxn=50;
int L_dfs ();
int R_dfs ();
int S_bfs (); int W,H,X1,Y1,X2,Y2,Q[MAXN*MAXN],FA[MAXN][MAXN], dlx[4]={0,-1,0,1},dly[4]={-1,0,1,0}, drx[4]={0,-1,0,1},dry[4]={
1,0,-1,0};
Char map[maxn][maxn],vis[maxn][maxn]={{0}};
	int main () {//Freopen ("In.txt", "R", stdin);
	int t,i,j;
	scanf ("%d", &t);
		while (t--) {scanf ("%d%d", &w,&h);
			for (i=0;i 
 
   This is its own version and feels a little improved. 

#include <stdio.h> #include <memory.h> const int maxn=50;
int Lr_dfs (int kg);
int BFS ();
int W,H,X1,Y1,X2,Y2,Q[MAXN*MAXN],DIS[MAXN*MAXN],VIS[MAXN*MAXN];
int FA[MAXN][MAXN],VIS[MAXN][MAXN];	int dlx[4]={0,-1,0,1},dly[4]={-1,0,1,0}, drx[4]={0,-1,0,1},dry[4]={1,0,-1,0};
Dlx[i]=drx[i], dly[i]=dry[(i+2)%4];
Char MAP[MAXN][MAXN];
	int main () {//Freopen ("In.txt", "R", stdin);
	int t,i,j;
	scanf ("%d", &t);
		while (t--) {scanf ("%d%d", &w,&h);
			for (i=0;i 

Summarize:

0
/ | \
1 2 3
/ | | \   \
4 5 6 7 8

(BFS Chart)
Issues to consider.
When ①bfs, prevent the walk, need a vis array mark, and in the judgment to avoid short-circuit operator interference;
② Search, find the exit after the need to find, through the global variable OK to control, especially in the DFS;
③ more use printf debugging intermediate process to find the cause of the error;