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  • --hdu2717->搜索的范围控制

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

    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. 

    之所以把这个水题贴上,是为了警戒自己,数组超限与范围判断的时候,应该先进行范围判断,然后进行数组操作!

    #include<stdio.h>
    #include<string.h>
    #include<queue>
    #include<algorithm>
    using namespace std;
    int vis[300010];
    
    int bfs(int x,int ans){
    
        queue<int>q;
        queue<int>num;
        q.push(x);
        num.push(0);
        while(!q.empty()){
    
            int xx = q.front();
            int nn = num.front();
            num.pop();
            q.pop();
    
            if(xx == ans)return nn;
            if(xx>=0&&vis[xx-1] == 0){
                q.push(xx-1);
                num.push(nn+1);
                vis[xx-1] = 1;
            }
            if(xx+1<=ans&&vis[xx+1] == 0){
                q.push(xx+1);
                num.push(nn+1);
                vis[xx+1] = 1;
            }
    
            if(2*xx <=ans*2&&vis[2*xx] == 0){
                q.push(2*xx);
                num.push(nn+1);
                vis[2*xx] = 1;
            }
    
        }
    
    }
    
    int main(){
    
    
        int n,m;
    
        while(~scanf("%d%d",&n,&m)){
             memset(vis,0,sizeof(vis));
             vis[n]=1;
             printf("%d
    ",bfs(n,m));
    
        }
    }
    

      

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  • 原文地址:https://www.cnblogs.com/lovelystone/p/4713229.html
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