UVA - 532 Dungeon Master

来源：互联网发布：excel2007数据库驱动编辑：程序博客网时间：2024/06/08 08:08

题目大意：逃出地牢。L 层长 R 宽 C，从 S 走到 E 要移动几次（秒），无法逃出输出 Trapped!

解题思路：. 可走，# 不能走，化为 0 1，0 可走 1 不可走，走过的位置标记为 1。仿照上一题移动骑士UVA - 439 Knight Moves ，移动方式不一样而已，可以在三维平面上下层移动，坐标移动方式改一下，感觉没太大区别。

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>using namespace std;int map[50][50][50];char tmp[50][50][50];int a, b, c;int L, R, C;struct queue {    int x, y, z;    int t;    queue(int a = 0, int b = 0, int c = 0, int d = 0):x(a),y(b),z(c),t(d){}};queue q[100000];int dd[6][3] = {{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,-1},{0,0,1}};int bfs(int x, int y, int z) {    if (x == a && y == b && z == c) return 0;    q[0] = queue(x, y, z, 0);    int t1 = 0, t2 = 1;    while (t1 < t2) {        struct queue now = q[t1++];        for (int i = 0; i < 6; i++) {            x = now.x + dd[i][0];            y = now.y + dd[i][1];            z = now.z + dd[i][2];            if (x < 0 || x >= L || y < 0 || y >= R || z < 0 || z >= C) continue;            if (!map[x][y][z]) {                if (x == a && y == b && z == c) return now.t+1;                q[t2++] = queue(x, y, z, now.t+1);                map[x][y][z] = 1;            }        }    }    return -1;}int main() {    while (scanf("%d%d%d", &L, &R, &C) != EOF && L+R+C) {        memset(map, 0, sizeof(map));        int x, y, z;        for (int i = 0; i < L; i++)            for (int j = 0; j < R; j++)                scanf("%s", tmp[i][j]);        for (int i = 0; i < L; i++)            for (int j = 0; j < R; j++)                for (int k = 0; k < C; k++)                    if (tmp[i][j][k] == 'S') x = i, y = j, z = k;                    else if (tmp[i][j][k] == 'E') a = i, b = j, c = k;                    else if (tmp[i][j][k] == '#') map[i][j][k] = 1;        int ans = bfs(x, y, z);        if (ans == -1) printf("Trapped!\n");        else printf("Escaped in %d minute(s).\n", ans);    }return 0;}

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