初识BFS POJ-3278 Catch That Cow FZU-2285 迷宫寻宝

令人窒息的创新实验课让我们写程序。
POJ-3278 Catch That Cow

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Input

Line 1: Two space-separated integers: N and K

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Output
```
5 17
```

Hint

The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.

我的题解

#include <iostream>
#include <queue>
using namespace std;
#define maxn 100001int step[maxn];   //代表到达每个位置需要的步数
bool visit[maxn]; //代表某个位置是否被访问
queue<int> q;
int BFS(int N, int K)
{int head, next;step[N] = 0;visit[N] = true;q.push(N);while (!q.empty()){head = q.front();q.pop();for (int i = 0; i < 3; i++){if (i == 0) //代表往左1next = head - 1;if (i == 1) //代表往右1next = head + 1;if (i == 2) //代表乘2next = head * 2;if (next < 0 || next > maxn) //排除越界的情况continue;if (!visit[next]){q.push(next); //加入队列中，下次作为head弹出，准备下一轮搜索visit[next] = true;step[next] = step[head] + 1; //head表示前一个点，next表示后一个点，两者的步数当然差1}if (next == K)return step[next];}}return 0; //代表找不到（大概，应该都会找到吧
}
int main()
{int N, K; //N是John初始的位置，K是牛的位置cin >> N >> K;if (N - K >= 0)cout << N - K << endl;elsecout << BFS(N, K) << endl;return 0;
}

然后是第二题 FZU-2285

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。每组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
```

我的题解

#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
#define maxn 1000bool visit[maxn][maxn];
int step[maxn][maxn];typedef struct
{int i; //代表节点的横坐标int j; //纵坐标
} node;queue<node> q; //定义一个节点为node的队列int BFS(char **maze, int n, node start, node end)
{visit[start.i][start.j] = true;step[start.i][start.j] = 0;q.push(start);   //关于起点的初始化node head, next; //在过程中分别指代现在的位置和下一个位置while (!q.empty()){head = q.front();q.pop();for (int i = 0; i < 4; i++){if (i == 0) //代表往左next.i = head.i - 1, next.j = head.j;if (i == 1) //代表往右next.i = head.i + 1, next.j = head.j;if (i == 2) //代表往上next.i = head.i, next.j = head.j - 1;if (i == 3) //代表往下next.i = head.i, next.j = head.j + 1;if (next.i < 0 || next.i >= n || next.j < 0 || next.j >= n || maze[next.i][next.j] == '#') //越界或者碰到墙壁的情况排除continue;if (!visit[next.i][next.j]){visit[next.i][next.j] = true;step[next.i][next.j] = step[head.i][head.j] + 1; //next的步数比head的步数+1q.push(next);}if (maze[next.i][next.j] == 'E')return step[next.i][next.j];}}return -1;
}
int main()
{int n; //表示迷宫的大小while (cin >> n){memset(visit, false, sizeof(visit));memset(step, 0, sizeof(step));while (!q.empty())q.pop();char **maze = new char *[n]; //动态创建二维数组mazefor (int i = 0; i < n; i++)maze[i] = new char[n];node start, end; //分别代表起点和终点，用来储存他们的坐标for (int i = 0; i < n; i++)scanf("%s", maze[i]);for (int i = 0; i < n; i++)for (int j = 0; j < n; j++){if (maze[i][j] == 'S')start.i = i, start.j = j;if (maze[i][j] == 'E')end.i = i, end.j = j;}cout << BFS(maze, n, start, end) << endl;}return 0;
}

水博客，逃(

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