HDU2717 Catch That Cow ( BFS )

Catch That Cow

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

 

Description

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 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?

 

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Input

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 Line 1: Two space-separated integers: N and K

 

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Output

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Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

 

 

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Sample Input

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5 17

 

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Sample Output

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 4

 

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Hint

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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. 

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 Analysis

 

　　从一个点到另一个点，每一步有多种走法，这是典型的 bfs 问题。

 

 

Tips

•  注意数组开的够不够大
•  在每一次开始判断是否已经搜到解

#include <stdio.h>
#include <string.h>
#include<queue>

using namespace std;

int vis[120000];
int n,k,flag;

struct Node
{
    int x;
    int tm;
}root;

void bfs()
{
    memset(vis,0,sizeof(vis));
    queue<Node>que;
    que.push(root);

    vis[root.x] = 1;
    int ans=100005;

    while(!que.empty())
    {
        Node q = que.front();
        que.pop();

        //printf("q.x = %d
",q.x);

        if(q.x == k )
        {
            ans = q.tm;
            flag = 1;
            break;
        }

        Node q1,q2,q3;

        if(q.x+1 < 100005 &&!vis[q.x + 1]   )
        {
            //printf("q2.x = %d
",q.x+1);
            q2.x = q.x + 1;
            q2.tm = q.tm + 1;

            vis[q2.x] = 1;
            que.push(q2);
        }

        if(q.x-1 >=0 && !vis[q.x - 1]   )
        {
            //printf("q1.x = %d
",q.x-1);
            q1.x = q.x - 1;
            q1.tm = q.tm + 1;

            vis[q1.x] = 1;
            que.push(q1);
        }



        if(q.x *2 <100005 && !vis[q.x *2]   )
        {
            //printf("q3.x = %d
",q.x*2);
            q3.x = q.x *2;
            q3.tm = q.tm + 1;

            vis[q3.x] = 1;
            que.push(q3);

        }
    }
    if(flag) printf("%d
",ans);
    //else printf("-1
");
}


int main()
{

    while(~scanf("%d%d",&n,&k))
    {
        flag = 0;

        root.x = n;
        root.tm = 0;

        bfs();
    }

    return 0;
}