C语言BFS(3)___Catch That Cow(Hdu 2717)

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

题目大概意思：在一条道上，每个点x可以一步到 x-1 , x+1 , 2*x 这三个点,输入两个点,问最短几步可以到达.

简单的BFS,除了队列大小开合适外其他没什么要注意的.

#include <stdio.h>#include <string.h>struct Team{    int x,s;}team[1000000];                  //两个数组不要开小了int book[200000];int main(){    int flag,left,right;    int head,tail;    while(scanf("%d%d",&left,&right)!=EOF)    {        flag=0;        if(left==right)        {            printf("0\n");            continue;        }        if(left>right)        {            printf("%d\n",left-right);            continue;        }        memset(book,0,sizeof(book));        head=tail=0;        team[tail].s=0;        team[tail++].x=left;        book[left]=1;        int m=right-left;        while(head<tail)        {                        if(team[head].s>m)           //剪枝            {                head++;                continue;            }            if(team[head].x>1&&!book[team[head].x-1])            {                team[tail].s=team[head].s+1;                team[tail++].x=team[head].x-1;                book[team[head].x-1]=1;            }            if(team[head].x<200000&&!book[team[head].x+1])            {                team[tail].s=team[head].s+1;                team[tail++].x=team[head].x+1;                book[team[head].x+1]=1;            }            if(team[head].x<100000&&!book[2*team[head].x])            {                team[tail].s=team[head].s+1;                team[tail++].x=2*team[head].x;                book[2*team[head].x]=1;            }            if(team[head].x-1==right||team[head].x+1==right||team[head].x*2==right)            {                printf("%d\n",team[head].s+1);                flag=1;                break;            }            head++;        }    }    return 0;}