HDU 2717 Catch That Cow (深搜)
Problem 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
HintThe 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.
分析:
每次走的话可以有三种走法,-1,+1,*2
代码:
#include<iostream>
#include<stdio.h>
#include<string>
#include<cstring>
#include<math.h>
#include<queue>
using namespace std;
int N=1000000;
int a[1010000]={0};防止走重,即第一步走过了后面不应该再走一次
struct ft//结构体,包含数和步数
{
int x,y;
}p;
int k;
int bfs(int n)
{
queue<ft>Q;//建队
p.x=n,p.y=0;
a[n]=1;
Q.push(p);入队
ft cur, nex;
while(!Q.empty())
{
cur=Q.front();取队首元素
Q.pop();出队
if(cur.x==k)return cur.y;//判断是否达到目的
nex.x=cur.x+1;//判断+1的操作是否合法
if(nex.x>=0&&nex.x<N&&a[nex.x]==0)
{
nex.y=cur.y+1;
a[nex.x]=1;
Q.push(nex);合法则入队并步数+1
}
nex.x=cur.x-1;
if(nex.x>=0&&nex.x<N&&a[nex.x]==0)
{
nex.y=cur.y+1;
a[nex.x]=1;
Q.push(nex);
}
nex.x=cur.x*2;
if(nex.x>=0&&nex.x<N&&a[nex.x]==0)
{
nex.y=cur.y+1;
a[nex.x]=1;
Q.push(nex);
}
}
return -1;
}
int main()
{
int n;
while(~scanf("%d %d",&n,&k))
{
memset(a,0,sizeof(a));//多次运行,刷0;
int bs=bfs(n);
if(bs>=0)
printf("%d\n",bs);
}
return 0;
}
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