POJ 2251 Dungeon Master BFS
2012-02-12 19:57
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这题其实是一题简单的BFS,代码注释如下程序所示:
/* ----------------------------------------------------- Time : 19:00 - 19:53 2012.2.12 stratege : BFS Author : Johnsondu ----------------------------------------------------- Problem: 2251 User: a312745658 Memory: 808K Time: 32MS Language: G++ Result: Accepted ----------------------------------------------------- */ #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <math.h> #include <queue> using namespace std ; int direct_x[6] = {1, -1, 0, 0, 0, 0} ; //6个方向,分别为东南西北上下 int direct_y[6] = {0, 0, 1, -1, 0, 0} ; //方向很好确定,可以看到没对应一次移动 int direct_z[6] = {0, 0, 0, 0, 1, -1} ; //只有一个坐标发生变化 char map[35][35][35] ; bool mark[35][35][35] ; //标记已走过的路程 int L, R, C ; bool flag ; int sx, sy, dx, dy, sz, dz ; //确定起点和终点坐标 struct Node { int x ; int y ; int z ; int step ; }n, m; int main() { int i, j, k ; while (cin >> L >> R >> C) { if (L == 0 && R == 0 && C == 0) break ; for (i = 1; i <= L; i ++) for (j = 1; j <= R ; j ++) for (k = 1; k <= C; k ++) { cin >> map[i][j][k] ; if (map[i][j][k] == 'S') sx = j, sy = k, sz = i ; if (map[i][j][k] == 'E') dz = i, dx = j, dy = k ; } flag = false ; n.z = sz ; //初始化结构体,进队列 n.x = sx ; n.y = sy ; n.step = 0 ; memset (mark, false, sizeof(mark)) ; queue <Node> Q ; Q.push (n) ; mark[n.z][n.x][n.y] = true ; while (! Q.empty()) { m = Q.front() ; Q.pop() ; if (map[m.z][m.x][m.y] == 'E') //寻找到目标 { flag = true ; break ; } for (i = 0; i < 6; i ++) { n.x = m.x + direct_x[i] ; n.y = m.y + direct_y[i] ; n.z = m.z + direct_z[i] ; if (n.x > 0 && n.y > 0 && n.x <= R && n.y <= C && map[n.z][n.x][n.y] != '#' && n.z > 0 && n.z <= L && mark[n.z][n.x][n.y] == false) { //当且仅当满足上列条件,才允许进队列 n.step = m.step + 1 ; //满足条件,步数加1 mark[n.z][n.x][n.y] = true ; //标记已走过 Q.push (n) ; } } } if (flag) printf ("Escaped in %d minute(s).\n", m.step) ; else printf ("Trapped!\n") ; } return 0 ; }
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