福州大学第十五届程序设计竞赛_重现赛B题迷宫寻宝
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
5
S.#..
#.#.#
#.#.#
#...E
#....
Sample Output
7
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）
#include <iostream>
#include <algorithm>
#include <vector>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <queue>
using namespace std; #define FF(i, a, b) for(int i = a; i < b; i++)
#define RR(i, a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define PR(x) printf("%d\n", x)
#define INF 0x3f3f3f3f
#define MAX 1100
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP; struct Node{
int x, y, step;
};
Node bg, ed, p1, p2;
queue<Node> q; char matrix[MAX][MAX];
int dir[][] = {,,-,,,,,-};
int dis[MAX][MAX];
int n, m, ans; int check(int x, int y){
if(x < || x >= n || y < || y >= m || dis[x][y] || matrix[x][y] == '#') return ;
return ;
}
void Clear(queue<Node>& q){
queue<Node> empty;
swap(empty, q);
} int BFS(){
p1 = bg;
q.push(p1);dis[p1.x][p1.y] = ;
while(!q.empty()){
p2 = q.front();q.pop();
if(p2.x == ed.x && p2.y == ed.y) return p2.step; //Í˳ö
for(int i = ; i < ; i++){
Node p3;
p3.x = p2.x + dir[i][];
p3.y = p2.y + dir[i][];
p3.step = p2.step;
if(!check(p3.x, p3.y)) continue;
dis[p3.x][p3.y] = ;
p3.step = p2.step + ;
q.push(p3);
}
}
return -;
} int main(void){
// #ifdef LOCAL
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
// #endif
ios::sync_with_stdio(false); cin.tie();
int i, j, k; while(cin>>n){
m = n;
Clear(q);ME(dis, ); for(i = ; i < n; i++){
for(j = ; j < m; j++){
cin>>matrix[i][j];
if(matrix[i][j] == 'S'){bg.x = i; bg.y = j; bg.step = ;}
else if(matrix[i][j] == 'E'){ed.x = i; ed.y = j;}
}
} ans = BFS();
cout<<ans<<endl;
}
return EXIT_SUCCESS;
}
POJ2251：三维空间找'S'到‘E’的距离，差不多的解法。
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. **/
#include <iostream>
#include <algorithm>
#include <vector>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <queue>
using namespace std;
#define FF(i, a, b) for(int i = a; i < b; i++)
#define RR(i, a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define PR(x) printf("%d\n", x)
#define INF 0x3f3f3f3f
#define MAX 35
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP;
struct Node{
int x, y, z, step;
};
Node bg, ed, p1, p2;
queue<Node> q;
char matrix[MAX][MAX][MAX];
int dir[][] = {,,,-,,,,,,,-,,,,,,,-};
int dis[MAX][MAX][MAX];
int n, m, l, ans;
int check(int x, int y, int z){
if(x < || x >= n || y < || y >= m || z < || z >= l || dis[x][y][z] || matrix[x][y][z] == '#') return ;
return ;
}
void Clear(queue<Node>& q){
queue<Node> empty;
swap(empty, q);
}
int BFS(){
p1 = bg;
q.push(p1);dis[p1.x][p1.y][p1.z] = ;
while(!q.empty()){
p2 = q.front();q.pop();
if(p2.x == ed.x && p2.y == ed.y && p2.z == ed.z) return p2.step; //退出
for(int i = ; i < ; i++){
Node p3;
p3.x = p2.x + dir[i][];
p3.y = p2.y + dir[i][];
p3.z = p2.z + dir[i][];
p3.step = p2.step;
if(!check(p3.x, p3.y, p3.z)) continue;
dis[p3.x][p3.y][p3.z] = ;
p3.step = p2.step + ;
q.push(p3);
}
}
return ;
}
int main(void){
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
ios::sync_with_stdio(false); cin.tie();
int i, j, k;
while(cin>>l>>n>>m){
if(l <= || n <= || m <= ) break;
Clear(q);ME(dis, );
for(k = ; k < l; k++){
for(i = ; i < n; i++){
for(j = ; j < m; j++){
cin>>matrix[i][j][k];
if(matrix[i][j][k] == 'S'){bg.x = i; bg.y = j; bg.z = k; bg.step = ;}
else if(matrix[i][j][k] == 'E'){ed.x = i; ed.y = j; ed.z = k;}
}
}
}
ans = BFS();
if(ans) cout<<"Escaped in "<<ans<<" minute(s)."<<endl;
else cout<<"Trapped!"<<endl;
}
return EXIT_SUCCESS;
}
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