HDU2717 Catch That Cow

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12690 Accepted Submission(s): 3924

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

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.

题意：
给你一个数n，一个数k，问通过n+1或n-1或n*2这三种操作最少能通过几次操作得到k

思路：
广搜这三种操作即可
要注意两个地方：
1，搜过的数字进行标记，不要重复搜索，不然会爆内存
2，特判下初始n==k的情况==

代码：

#include<cstdio>#include<iostream>#include<cstring>#include<queue>using namespace std;int N=1000000;int map[1000010];struct node{    int value;    int step;};int main(){    int n;    int k;    while(scanf("%d%d",&n,&k)!=EOF)    {        memset(map,0,sizeof(map));        if(n==k)        {            printf("0\n");            continue;        }        queue<node>q;        node x;        x.value=n;        x.step=0;        q.push(x);        while(!q.empty())        {            node item;            item.value=(q.front().value+1);            if(item.value>=0&&item.value<=N&&map[item.value]==0)            {                map[item.value]=1;                item.step=(q.front().step+1);                q.push(item);                if(item.value==k)                break;            }            item.value=(q.front().value-1);            if(item.value>=0&&item.value<=N&&map[item.value]==0)            {                map[item.value]=1;                item.step=(q.front().step+1);                q.push(item);                if(item.value==k)                break;            }            item.value=(q.front().value*2);            if(item.value>=0&&item.value<=N&&map[item.value]==0)            {                map[item.value]=1;                item.step=(q.front().step+1);                q.push(item);                if(item.value==k)                break;            }            q.pop();        }        printf("%d\n",q.back().step);    }    return 0;}