【搜索入门专题1】hdu2717 H

Catch That Cow

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
 

题意：输入起始点和终止点，输出起始点到终止点的步数，从起始点到终止点有三种走法，1：自身步数+1；2：自身步数-1；3：自身步数*2。

思路：从起始点开始，搜索三种走法能到达的点，满足条件的点加入队列，直到找到终止点，结束函数，输出步数。

易错点：判断该点是否走过之前，首先该点必须满足在数组范围内，RE+MEL  11遍  最后还是师父给指点迷津

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<queue>
using namespace std;
#define N 1001000
int book[N];
int startx,endx;
int sumtime;

struct note{
	int x;
	int time;
};

void BFS()
{
	note head,now_head,next1,next2,next3;
	queue<note>Q;
	head.time = 0;
	head.x = startx;
	book[startx] = 1;
	Q.push(head);
	while(!Q.empty())
	{
		now_head = Q.front() ;
		Q.pop() ;
		if(now_head.x == endx)//找到终止点，结束函数 
		{
			sumtime = now_head.time ;
			return;
		}
		next1.x  = now_head.x + 1;//向前走一步 
		next2.x  = now_head.x - 1;//往后退一步 
		next3.x  = now_head.x + now_head.x;//自身两倍 
		//注意步数范围
		//注意先后顺序，首先满足在范围内，再判断是否走过 
		if(next1.x < N&&next1.x>=0&&!book[next1.x])
		{
			book[next1.x]  = 1;
			next1.time = now_head.time + 1;
			Q.push(next1); 
		}
		
		if(next2.x < N&&next2.x >= 0&&!book[next2.x])
		{
			book[next2.x]  = 1;
			next2.time = now_head.time + 1;
			Q.push(next2); 
		}
		
		if(next3.x < N&&next3.x >= 0&&!book[next3.x])
		{
			book[next3.x]  = 1;
			next3.time = now_head.time + 1;
			Q.push(next3); 
		}
	}
	return;
}

int main()
{
	while(scanf("%d%d",&startx,&endx)!=EOF)
	{
		sumtime = 0;
		memset(book,0,sizeof(book));
		BFS();
		printf("%d
",sumtime);
	}
	return 0;
}