福州大学第十五届程序设计竞赛_重现赛B题迷宫寻宝

时间：2018-05-14 21:37:17 阅读：401 评论：0 收藏：0 [点我收藏+]

标签：mes 三维空间 node rap turn dad bmp swap col

Problem B 迷宫寻宝

Accept: 52 Submit: 183
Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem Description

洪尼玛今天准备去寻宝，在一个n*n （n行, n列）的迷宫中，存在着一个入口、一些墙壁以及一个宝藏。由于迷宫是四连通的，即在迷宫中的一个位置，只能走到与它直接相邻的其他四个位置（上、下、左、右）。现洪尼玛在迷宫的入口处，问他最少需要走几步才能拿到宝藏？若永远无法拿到宝藏，则输出-1。

Input

多组测试数据。

输入第一行为正整数n，表示迷宫大小。

接下来n行，每行包括n个字符，其中字符‘.‘表示该位置为空地，字符‘#‘表示该位置为墙壁，字符‘S‘表示该位置为入口，字符‘E‘表示该位置为宝藏，输入数据中只有这四种字符，并且‘S‘和‘E‘仅出现一次。

n≤1000

Output

输出拿到宝藏最少需要走的步数，若永远无法拿到宝藏，则输出-1。

Sample Input

S.#..

#.#.#

#...E

#....

Sample Output

分析：DFS，BFS基础题型，下面加一个poj2251，加深对简单搜索的理解，三维空间是3维数组，如果加上钥匙或者传送门的话就是四维空间

详情见: 2018年长沙理工大学G题（找钥匙） 2018年湘潭大学F题（传送门）

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <vector>
 4 #include <stdlib.h>
 5 #include <string.h>
 6 #include <math.h>
 7 #include <queue>
 8 using namespace std;
 9 
10 #define FF(i, a, b) for(int i = a; i < b; i++)
11 #define RR(i, a, b) for(int i = a; i > b; i++)
12 #define ME(a, b) memset(a, b, sizeof(a))
13 #define SC(x) scanf("%d", &x)
14 #define PR(x) printf("%d\n", x)
15 #define INF 0x3f3f3f3f
16 #define MAX 1100
17 #define MOD 1000000007
18 #define E 2.71828182845
19 #define M 8
20 #define N 6
21 typedef long long LL;
22 const double PI = acos(-1.0);
23 typedef pair<int, int> Author;
24 vector<pair<string, int> > VP; 
25 
26 struct Node{
27     int x, y, step;
28 };
29 Node bg, ed, p1, p2;
30 queue<Node> q;
31 
32 char matrix[MAX][MAX];
33 int dir[4][2] = {1,0,-1,0,0,1,0,-1};
34 int dis[MAX][MAX];
35 int n, m, ans;
36 
37 int check(int x, int y){
38     if(x < 0 || x >= n || y < 0 || y >= m || dis[x][y] || matrix[x][y] == ‘#‘) return 0;
39     return 1;
40 }
41 void Clear(queue<Node>& q){
42     queue<Node> empty;
43     swap(empty, q);
44 }
45 
46 int BFS(){
47     p1 = bg;
48     q.push(p1);dis[p1.x][p1.y] = 1;
49     while(!q.empty()){
50         p2 = q.front();q.pop();
51         if(p2.x == ed.x && p2.y == ed.y) return p2.step;        //ÍË³ö 
52         for(int i = 0; i < 4; i++){
53             Node p3;
54             p3.x = p2.x + dir[i][0];
55             p3.y = p2.y + dir[i][1];
56             p3.step = p2.step;
57             if(!check(p3.x, p3.y)) continue;
58             dis[p3.x][p3.y] = 1;
59             p3.step = p2.step + 1;
60              q.push(p3);
61         }
62     }
63     return -1;
64 }
65 
66 int main(void){
67 //    #ifdef LOCAL
68 //        freopen("in.txt", "r", stdin);
69 //        freopen("out.txt", "w", stdout);
70 //    #endif
71     ios::sync_with_stdio(false); cin.tie(0);
72     int i, j, k;
73     
74     while(cin>>n){
75         m = n;
76         Clear(q);ME(dis, 0);
77         
78         for(i = 0; i < n; i++){
79             for(j = 0; j < m; j++){
80                 cin>>matrix[i][j];
81                 if(matrix[i][j] == ‘S‘){bg.x = i; bg.y = j; bg.step = 0;}
82                 else if(matrix[i][j] == ‘E‘){ed.x = i; ed.y = j;}
83             }
84         }
85         
86         ans = BFS();
87         cout<<ans<<endl;
88     }
89     return EXIT_SUCCESS;
90 }

View Code

POJ2251：三维空间找‘S‘到‘E’的距离，差不多的解法。

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 71                                                                         
 72 #include <iostream>
 73 #include <algorithm>
 74 #include <vector>
 75 #include <stdlib.h>
 76 #include <string.h>
 77 #include <math.h>
 78 #include <queue>
 79 using namespace std;
 80 
 81 #define FF(i, a, b) for(int i = a; i < b; i++)
 82 #define RR(i, a, b) for(int i = a; i > b; i++)
 83 #define ME(a, b) memset(a, b, sizeof(a))
 84 #define SC(x) scanf("%d", &x)
 85 #define PR(x) printf("%d\n", x)
 86 #define INF 0x3f3f3f3f
 87 #define MAX 35
 88 #define MOD 1000000007
 89 #define E 2.71828182845
 90 #define M 8
 91 #define N 6
 92 typedef long long LL;
 93 const double PI = acos(-1.0);
 94 typedef pair<int, int> Author;
 95 vector<pair<string, int> > VP; 
 96 
 97 struct Node{
 98     int x, y, z, step;
 99 };
100 Node bg, ed, p1, p2;
101 queue<Node> q;
102 
103 char matrix[MAX][MAX][MAX];
104 int dir[6][3] = {1,0,0,-1,0,0,0,1,0,0,-1,0,0,0,1,0,0,-1};
105 int dis[MAX][MAX][MAX];
106 int n, m, l, ans;
107 
108 int check(int x, int y, int z){
109     if(x < 0 || x >= n || y < 0 || y >= m || z < 0 || z >= l || dis[x][y][z] || matrix[x][y][z] == ‘#‘) return 0;
110     return 1;
111 }
112 void Clear(queue<Node>& q){
113     queue<Node> empty;
114     swap(empty, q);
115 }
116 
117 int BFS(){
118     p1 = bg;
119     q.push(p1);dis[p1.x][p1.y][p1.z] = 1;
120     while(!q.empty()){
121         p2 = q.front();q.pop();
122         if(p2.x == ed.x && p2.y == ed.y && p2.z == ed.z) return p2.step;        //退出 
123         for(int i = 0; i < 6; i++){
124             Node p3;
125             p3.x = p2.x + dir[i][0];
126             p3.y = p2.y + dir[i][1];
127             p3.z = p2.z + dir[i][2];
128             p3.step = p2.step;
129             if(!check(p3.x, p3.y, p3.z)) continue;
130             dis[p3.x][p3.y][p3.z] = 1;
131             p3.step = p2.step + 1;
132              q.push(p3);
133         }
134     }
135     return 0;
136 }
137 
138 int main(void){
139     #ifdef LOCAL
140         freopen("in.txt", "r", stdin);
141         freopen("out.txt", "w", stdout);
142     #endif
143     ios::sync_with_stdio(false); cin.tie(0);
144     int i, j, k;
145     
146     while(cin>>l>>n>>m){
147         if(l <= 0 || n <= 0 || m <= 0) break;
148         Clear(q);ME(dis, 0);
149         
150         for(k = 0; k < l; k++){
151             for(i = 0; i < n; i++){
152                 for(j = 0; j < m; j++){
153                     cin>>matrix[i][j][k];
154                     if(matrix[i][j][k] == ‘S‘){bg.x = i; bg.y = j; bg.z = k; bg.step = 0;}
155                     else if(matrix[i][j][k] == ‘E‘){ed.x = i; ed.y = j; ed.z = k;}
156                 }
157             }
158         }
159         
160         ans = BFS();
161         if(ans) cout<<"Escaped in "<<ans<<" minute(s)."<<endl;
162         else cout<<"Trapped!"<<endl;
163         
164     }
165     return EXIT_SUCCESS;
166 }