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题目大意: 给出一三维空间的地牢,要求求出由字符‘S‘到字符‘E‘的最短路径,移动方向可以是上,下,左,右,前,后,六个方向,每移动一次就耗费一分钟,要求输出最快的走出时间。不同L层的地图,相同RC坐标处是相连通的。(.可走,#为墙)
解题思路:从起点开始分别往6个方向进行BFS(即入队),并记录步数,直至队为空.若一直找不到,则困住。
/* POJ 2251 Dungeon Master --- 三维BFS(用BFS求最短路) */ #include <cstdio> #include <cstring> #include <queue> using namespace std; int n, m, o; //n行m列,o为层 char mapp[31][31][31]; bool visit[31][31][31]; //标记访问状态 /* 构造队列元素 (点位置+步数)*/ struct Point{ int x, y, z; int step; Point(){} Point(int a, int b, int c, int d) :x(a), y(b), z(c), step(d){} }; /* 判断点是否满足访问条件 */ inline bool judge(Point tmp){ if (mapp[tmp.x][tmp.y][tmp.z] == ‘#‘ || visit[tmp.x][tmp.y][tmp.z] || tmp.x < 0 || tmp.x >= o || tmp.y < 0 || tmp.y >= n || tmp.z < 0 || tmp.z >= m ){ return 0; } return 1; } /* x0为深度 y0,z0为同一层坐标 */ void bfs(int x0, int y0,int z0){ queue<Point> q; bool found = 0; visit[x0][y0][z0] = 1; //起点标记已访问 q.push(Point(x0, y0, z0, 0)); while (!q.empty()){ Point tmp = q.front(); q.pop(); if (mapp[tmp.x][tmp.y][tmp.z] == ‘E‘){ //找到终点,搜索结束,输出最短路 printf("Escaped in %d minute(s).\n", tmp.step); found = 1; break; } else{ Point tmp2; ++tmp.step; //步数必定增加 //往上一层 tmp2 = tmp; tmp2.x = tmp.x - 1; if (judge(tmp2)){ visit[tmp2.x][tmp2.y][tmp2.z] = 1; q.push(tmp2); } //往下一层 tmp2 = tmp; tmp2.x = tmp.x + 1; if (judge(tmp2)){ visit[tmp2.x][tmp2.y][tmp2.z] = 1; q.push(tmp2); } //以下四种为同一层 //往上一行 tmp2 = tmp; tmp2.y = tmp.y - 1; if (judge(tmp2)){ visit[tmp2.x][tmp2.y][tmp2.z] = 1; q.push(tmp2); } //往下一行 tmp2 = tmp; tmp2.y = tmp.y + 1; if (judge(tmp2)){ visit[tmp2.x][tmp2.y][tmp2.z] = 1; q.push(tmp2); } //往左一列 tmp2 = tmp; tmp2.z = tmp.z - 1; if (judge(tmp2)){ visit[tmp2.x][tmp2.y][tmp2.z] = 1; q.push(tmp2); } //往右一列 tmp2 = tmp; tmp2.z = tmp.z + 1; if (judge(tmp2)){ visit[tmp2.x][tmp2.y][tmp2.z] = 1; q.push(tmp2); } }//else }//while if (!found){ printf("Trapped!\n"); } } int main() { #ifdef _LOCAL freopen("D:\\input.txt", "r", stdin); #endif int sx, sy, sz; while (scanf("%d%d%d", &o, &n, &m) == 3 && (m + n + o)){ memset(visit, 0, sizeof visit); for (int i = 0; i < o; ++i){ for (int j = 0; j < n; ++j){ scanf("%s", mapp[i][j]); for (int k = 0; k < m; ++k){ if (mapp[i][j][k] == ‘S‘){ sx = i; sy = j; sz = k; } }//for(k) }//for(j) n行 }//for(i) o层 bfs(sx, sy, sz); //sx是深度 } return 0; }
POJ 2251 Dungeon Master --- 三维BFS(用BFS求最短路)
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原文地址:http://www.cnblogs.com/tommychok/p/5184498.html