UVa532 Dungeon Master 三维迷宫

 

 

学习点：

scanf可以自动过滤空行

搜索时要先判断是否越界（L R C），再判断其他条件是否满足

bfs搜索时可以在入口处（push时）判断是否达到目标，也可以在出口处（pop时）

 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<queue>
using namespace std;
const int N=31;
const int dl[]={-1, 1, 0, 0, 0, 0};
const int dr[]={0, 0, -1, 1, 0, 0};
const int dc[]={0, 0, 0, 0, -1, 1};

struct State {
    State(int l, int r, int c, int d): l(l), r(r), c(c), d(d) {}
    int l, r, c;
    int d;
};

int L, R, C;
int SL, SR, SC;
int EL, ER, EC;
typedef char A[N][N][N];
A dungeon, vis;

int bfs() {
    if(SL==EL && SR==EL && SC==EC) return 0;
    State s(SL, SR, SC, 0); vis[SL][SR][SC]=1;
    queue<State> q;
    q.push(s);
    while(!q.empty()) {
        State t=q.front(); q.pop();
        //可以在出口判断
        //if(t.l==EL && t.r==ER && t.c==EC) return t.d;
        for(int i=0;i<6;i++) {
            int nl=t.l+dl[i];
            int nr=t.r+dr[i];
            int nc=t.c+dc[i];
            if(nl>=0 && nl<L && nr>=0 && nr<R && nc>=0 && nc<C && !vis[nl][nr][nc] && dungeon[nl][nr][nc]!='#') {
                //也可以在入口判断
                if(nl==EL && nr==ER && nc==EC) return t.d+1;
                q.push(State(nl, nr, nc, t.d+1));
                vis[nl][nr][nc]=1;
            }
        }
    }
    return -1;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("./uva532.in", "r", stdin);
#endif
    while(scanf("%d%d%d", &L, &R, &C)==3 && L && R && C) {
        for(int i=0;i<L;i++) {
            for(int j=0;j<R;j++) {
                scanf("%s", dungeon[i][j]);
                for(int k=0;k<C;k++)
                    if(dungeon[i][j][k]=='S') SL=i, SR=j, SC=k;
                    else if(dungeon[i][j][k]=='E') EL=i, ER=j, EC=k;
            }
        }
        memset(vis, 0, sizeof vis);
        int ans=bfs();
        if(ans==-1)
            printf("Trapped!\n");
        else
            printf("Escaped in %d minute(s).\n", ans);
    }
    return 0;
}

 

 

// UVa532 Dungeon Master
// Rujia Liu
// 题意：三维迷宫中给定起点（字符S）和终点（字符E），墙是'#'，空格是'.'，求最短路长度
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;

struct State {
  int l, r, c;
  State(int l, int r, int c):l(l),r(r),c(c) {}
};

const int maxn = 30 + 5;
const int dl[] = {1,-1,0,0,0,0};
const int dr[] = {0,0,1,-1,0,0};
const int dc[] = {0,0,0,0,1,-1};
int L, R, C, d[maxn][maxn][maxn], vis[maxn][maxn][maxn];
char maze[maxn][maxn][maxn];

int bfs(int l1, int r1, int c1) {
  queue<State> Q;
  d[l1][r1][c1] = 0;
  vis[l1][r1][c1] = 1;
  Q.push(State(l1, r1, c1));
  while(!Q.empty()) {
    State s = Q.front(); Q.pop();
    for(int i = 0; i < 6; i++) {
      int newl = s.l + dl[i];
      int newr = s.r + dr[i];
      int newc = s.c + dc[i];
      if(newl >= 0 && newl < L && newr >= 0 && newr < R && newc >= 0 && newc < C && maze[newl][newr][newc] != '#' && !vis[newl][newr][newc]) {
        Q.push(State(newl, newr, newc));
        vis[newl][newr][newc] = 1;
        d[newl][newr][newc] = d[s.l][s.r][s.c] + 1;
        if(maze[newl][newr][newc] == 'E') return d[newl][newr][newc];
      }
    }
  }
  return -1;
}

int main() {
  while(scanf("%d%d%d", &L, &R, &C) == 3 && L) {
    int l1, r1, c1;
    for(int i = 0; i < L; i++)
      for(int j = 0; j < R; j++) {
        scanf("%s", maze[i][j]);
        for(int k = 0; k < C; k++)
          if(maze[i][j][k] == 'S') { l1 = i; r1 = j; c1 = k; }
      }
    memset(vis, 0, sizeof(vis));
    int ans = bfs(l1, r1, c1);
    if(ans >= 0) printf("Escaped in %d minute(s).\n", ans);
    else printf("Trapped!\n");
  }
  return 0;
}