POJ3278（BFS入门）

Problem Descrption

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 Input

5 17

Sample Output

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.

问题链接：http://poj.org/problem?id=3278

问题分析：搜索类问题，首先想到DFS和BFS两种，像这种很大的，如果用深搜很容易超时，所以明显是BFS，一开始写

为了记录步数使用了结构数组，然而这样会ML，这也给我提醒，以结构类型的队列可能内存较大，后来利用及数组记录父子节点的方式解决，可以建立数组ed[tx]=t，即记录tx的父节点为t，最后再回溯即可；

AC代码：

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
int n, k;
int vis[100005];
int ed[100005];
void bfs()
{queue<int>q;q.push(n);vis[n];while (!q.empty()){int t = q.front();if (t == k){return;}q.pop();int tx;for (int i = 0; i < 3; i++){if (i == 0) { tx = t + 1; }if (i == 1) { tx = t * 2; }if (i == 2) { tx = t - 1; }if (tx > 100000 || tx < 0 || vis[tx])continue;q.push(tx);ed[tx] = t;vis[tx]=1;}}
}
int main()
{while (cin >> n >> k){memset(vis, 0, sizeof(vis));bfs();int t=k,step=0;while (t!=n){t = ed[t];step++;}cout << step << endl;}
}

因使用结构队列而ML的代码：

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
int n, k,min1=999999;
int vis[100005];
struct way
{int x;int step;
};
void bfs()
{queue<way>q;struct way a;a.x = n, a.step = 0;q.push(a);vis[a.x];while (!q.empty()){struct way t = q.front();if (t.x == k){if (t.step < min1)min1 = t.step;return;}q.pop();struct way tx;for (int i = 0; i < 3; i++){if (i == 0) { tx.x = t.x + 1; tx.step =t.step+1; }if (i == 1) { tx.x = t.x * 2; tx.step =t.step+1; }if (i == 2) { tx.x = t.x - 1; tx.step = t.step + 1; }if (tx.x > 100000 || tx.x < 0 || vis[tx.x])continue;q.push(tx);vis[tx.x];}}
}
int main()
{while (cin >> n >> k){memset(vis, 0, sizeof(vis));min1 = 9999999;bfs();cout << min1 << endl;}
}

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