USACO: Overfencing

Overfencing
Kolstad and Schrijvers

Farmer John went crazy and created a huge maze of fences out in a field. Happily, he left out two fence segments on the edges, and thus created two "exits" for the maze. Even more happily, the maze he created by this overfencing experience is a `perfect' maze: you can find a way out of the maze from any point inside it.

Given W (1 <= W <= 38), the width of the maze; H (1 <= H <= 100), the height of the maze; 2*H+1 lines with width 2*W+1 characters that represent the maze in a format like that shown later - then calculate the number of steps required to exit the maze from the `worst' point in the maze (the point that is `farther' from either exit even when walking optimally to the closest exit). Of course, cows walk only parallel or perpendicular to the x-y axes; they do not walk on a diagonal. Each move to a new square counts as a single unit of distance (including the move "out" of the maze.

Here's what one particular W=5, H=3 maze looks like:

+-+-+-+-+-+
|         |
+-+ +-+ + +
|     | | |
+ +-+-+ + +
| |     |  
+-+ +-+-+-+

Fenceposts appear only in odd numbered rows and and odd numbered columns (as in the example). The format should be obvious and self explanatory. Each maze has exactly two blank walls on the outside for exiting.

PROGRAM NAME: maze1

INPUT FORMAT

Line 1:	W and H, space separated
Lines 2 through 2*H+2:	2*W+1 characters that represent the maze

SAMPLE INPUT (file maze1.in)

5 3
+-+-+-+-+-+
|         |
+-+ +-+ + +
|     | | |
+ +-+-+ + +
| |     |  
+-+ +-+-+-+

OUTPUT FORMAT

A single integer on a single output line. The integer specifies the minimal number of steps that guarantee a cow can exit the maze from any possible point inside the maze.

SAMPLE OUTPUT (file maze1.out)

9

The lower left-hand corner is *nine* steps from the closest exit.

 

做这道题和做上一道（脱离地牢）的时候学会了这种迷宫寻最短路的思路，类似填数游戏：起点填1，1的相邻的四格符合条件的填2，每个填2的相邻四格符合条件的填3……直到获得所需答案

其实就是传说中的BFS

/* ID: geterns1 PROG: maze1 LANG: C++ */ #include <iostream> #include <fstream> #include <cstring> using namespace std; int main(){ ifstream ifs("maze1.in"); ofstream ofs("maze1.out"); char map[201][77]; short w, h; bool stp[100][38] = { false }; short curStp = 0, beg = 0, counter = 0, rBeg, rCounter, x[5000], y[5000]; ifs >> w >> h; ifs.ignore(); for (short i = 0; i < 2 * h + 1; i ++) ifs.getline(map[i], 2 * w + 2); //find the exit for (short i = 0; i < h; i ++){ if (map[2 * i + 1][0] == ' '){ stp[i][0] = true; x[beg + counter] = i; y[beg + counter] = 0; counter ++; } if (map[2 * i + 1][2 * w] == ' '){ stp[i][w - 1] = true; x[beg + counter] = i; y[beg + counter] = w - 1; counter ++; } } for (short i = 0; i < w; i ++){ if (map[0][2 * i + 1] == ' '){ stp[0][i] = true; x[beg + counter] = 0; y[beg + counter] = i; counter ++; } if (map[2 * h][2 * i + 1] == ' '){ stp[h - 1][i] = true; x[beg + counter] = h - 1; y[beg + counter] = i; counter ++; } } while (counter > 0){ curStp ++; rBeg = beg + counter; rCounter = 0; for (short i = beg; i < beg + counter; i ++){ if (x[i] > 0 && map[2 * x[i]][2 * y[i] + 1] != '-' && stp[x[i] - 1][y[i]] == false){ stp[x[i] - 1][y[i]] = true; x[rBeg + rCounter] = x[i] - 1; y[rBeg + rCounter] = y[i]; rCounter ++; } if (x[i] < h - 1 && map[2 * x[i] + 2][2 * y[i] + 1] != '-' && stp[x[i] + 1][y[i]] == false){ stp[x[i] + 1][y[i]] = true; x[rBeg + rCounter] = x[i] + 1; y[rBeg + rCounter] = y[i]; rCounter ++; } if (y[i] > 0 && map[2 * x[i] + 1][2 * y[i]] != '|' && stp[x[i]][y[i] - 1] == false){ stp[x[i]][y[i] - 1] = true; x[rBeg + rCounter] = x[i]; y[rBeg + rCounter] = y[i] - 1; rCounter ++; } if (y[i] < w - 1 && map[2 * x[i] + 1][2 * y[i] + 2] != '|' && stp[x[i]][y[i] + 1] == false){ stp[x[i]][y[i] + 1] = true; x[rBeg + rCounter] = x[i]; y[rBeg + rCounter] = y[i] + 1; rCounter ++; } } beg = rBeg; counter = rCounter; } ofs << curStp << endl; ifs.close(); ofs.close(); return 0; }