题目链接：http://poj.org/problem?id=3278
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 47168 | Accepted: 14818 |
Description
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
Output
Sample Input
5 17
Sample Output
4
Hint
Source
思路：广搜入门题;（1）广搜运用一个队列让所有可行节点入队，然后再一个个出队判最优解；
（2）这道题用数组模拟队列~~~=。=
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <cstdio>
#include <cmath>
const int maxn=101000;
using namespace std;
int vis[maxn*2],que[maxn*2],d[maxn*2];
int n,k;
void bfs(int x)
{
memset(vis,0,sizeof(vis));
memset(que,0,sizeof(que));
memset(d,0,sizeof(d));
vis[x]=1;
int front=0,rear=0;
que[rear++]=x;
while(front<rear)
{
x=que[front++];
if(x==k)return; //广搜出口
//下面剪支
if(!vis[x-1]&&x>0) {vis[x-1]=1;que[rear++]=x-1;d[x-1]=d[x]+1;}
if(!vis[x+1]&&x<100000) {vis[x+1]=1;que[rear++]=x+1;d[x+1]=d[x]+1;}
if(!vis[2*x]&&2*x>=0&&2*x<=100000) {vis[2*x]=1;que[rear++]=x*2;d[x*2]=d[x]+1;}
}
}
int main()
{
while(cin>>n>>k)
{
bfs(n);
cout<<d[k]<<endl;
}
return 0;
}
原文地址：http://blog.csdn.net/liusuangeng/article/details/38959051
