POJ 2251 Dungeon Master（BFS + A*）
题意：
3维迷宫，求从起点到终点最少要走的时间，若不能走到，则输出“Trapped!”
思路：
1. 求最短路径首先想到 BFS，本题稍有变化就是在于 3 维迷宫，其实和 2 维迷宫都是一样的；
2. 解法一采用了优先队列 + 估值函数，但是看来时间上并没有优化太多，同样都是 16ms；
解法一：BFS + A*（16ms）
#include <iostream> #include <algorithm> #include <queue> using namespace std; const int MAXN = 35; const int dir[6][3] = {{1, 0, 0}, {-1, 0, 0}, {0, 1, 0}, {0, -1, 0}, {0, 0, 1}, {0, 0, -1}}; char gird[MAXN][MAXN][MAXN]; int L, R, C, ex, ey, ez; bool vis[MAXN][MAXN][MAXN]; struct ST { int x, y, z, step, eval; ST(int _x, int _y, int _z, int _s, int _e) : x(_x), y(_y), z(_z), step(_s), eval(_e) {} bool operator < (const ST& other) const { if (eval == other.eval) return step > other.step; return eval > other.eval; } }; bool judge(int x, int y, int z) { if (gird[z][x][y] != '#' && 0 < x && x <= R && 0 < y && y <= C && 0 < z && z <= L) { return true; } return false; } int evaluation(int x, int y, int z) { return abs(ex - x) + abs(ey - y) + abs(ez - z); } int bfs(int x, int y, int z) { priority_queue<ST> Q; Q.push(ST(x, y, z, 0, 0)); vis[z][x][y] = true; while (!Q.empty()) { ST s = Q.top(); Q.pop(); if (gird[s.z][s.x][s.y] == 'E') return s.step; for (int i = 0; i < 6; i++) { int a = s.x + dir[i][0]; int b = s.y + dir[i][1]; int c = s.z + dir[i][2]; if (judge(a, b, c) && !vis[c][a][b]) { vis[c][a][b] = true; int eval = s.step + 1 + evaluation(a, b, c); Q.push(ST(a, b, c, s.step + 1, eval)); } } } return -1; } int main() { while (scanf("%d%d%d", &L, &R, &C) && L && R && C) { int x, y, z; for (int i = 1; i <= L; i++) { for (int j = 1; j <= R; j++) { scanf("%s", &gird[i][j][1]); for (int k = 1; k <= C; k++) { if (gird[i][j][k] == 'S') x = j, y = k, z = i; if (gird[i][j][k] == 'E') ex = j, ey = k, ez = i; } } } memset(vis, false, sizeof(vis)); int ans = bfs(x, y, z); if (ans != -1) printf("Escaped in %d minute(s).\n", ans); else printf("Trapped!\n"); } return 0; }
解法二：普通 BFS（16ms）
#include <iostream>
#include <algorithm>
#include <deque>
using namespace std;
const int MAXN = 35;
const int dir[6][3] = {{1, 0, 0}, {-1, 0, 0}, {0, 1, 0}, {0, -1, 0}, {0, 0, 1}, {0, 0, -1}};
char gird[MAXN][MAXN][MAXN];
int L, R, C;
bool vis[MAXN][MAXN][MAXN];
struct ST {
int x, y, z, step;
ST(int _x, int _y, int _z, int _step) : x(_x), y(_y), z(_z), step(_step) {}
};
bool judge(int x, int y, int z) {
if (gird[z][x][y] != '#' &&
0 < x && x <= R && 0 < y && y <= C && 0 < z && z <= L) {
return true;
}
return false;
}
int bfs(int x, int y, int z) {
deque<ST> Q;
Q.push_back(ST(x, y, z, 0));
vis[z][x][y] = true;
while (!Q.empty()) {
ST s = Q.front();
Q.pop_front();
if (gird[s.z][s.x][s.y] == 'E')
return s.step;
for (int i = 0; i < 6; i++) {
int a = s.x + dir[i][0];
int b = s.y + dir[i][1];
int c = s.z + dir[i][2];
if (judge(a, b, c) && !vis[c][a][b]) {
vis[c][a][b] = true;
Q.push_back(ST(a, b, c, s.step + 1));
}
}
}
return -1;
}
int main() {
while (scanf("%d%d%d", &L, &R, &C) && L && R && C) {
int x, y, z;
for (int i = 1; i <= L; i++) {
for (int j = 1; j <= R; j++) {
scanf("%s", &gird[i][j][1]);
for (int k = 1; k <= C; k++) {
if (gird[i][j][k] == 'S')
x = j, y = k, z = i;
}
}
}
memset(vis, false, sizeof(vis));
int ans = bfs(x, y, z);
if (ans != -1)
printf("Escaped in %d minute(s).\n", ans);
else
printf("Trapped!\n");
}
return 0;
}