【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 1001000int 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\n",sumtime);}return 0;}