标签：des style blog color os 2014
<span style="color:#330099;">/*
A - 广搜 基础
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit
Status
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
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
4
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.
By Grant Yuan
2014.7.13
*/
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
#include<cstdio>
using namespace std;
bool a[100003];
typedef struct{
int num;
int sum;
}dd;
queue<dd>q;
int n,k;
int res;
void slove()
{int m,count;
int x;
dd init;
while(!q.empty()){
if(q.front().num==n)
{ res=q.front().sum;
break;
}
m=q.front().num;
count=q.front().sum;
x=m-1;
if(a[x]==0)
{ init.num=x;
init.sum=count+1;
q.push(init);
a[x]=1;
}
x=m+1;
if(a[x]==0)
{
init.num=x;
init.sum=count+1;
q.push(init);
a[x]=1;
}
if(m%2==0){
x=m/2;
if(a[x]==0)
{
init.num=x;
init.sum=count+1;
q.push(init);
a[x]=1;
}}
q.pop();}
}
int main()
{
scanf("%d%d",&n,&k);
memset(a,0,sizeof(a));
dd init;
init.num=k;
init.sum=0;
a[k]=1;
q.push(init);
slove();
cout<<res<<endl;
return 0;
}
</span>标签：des style blog color os 2014
原文地址：http://blog.csdn.net/yuanchang_best/article/details/37730423
