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  • 【题解】[Usaco2007 Open]Catch That Cow 抓住那只牛-C++

    题目
    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?

    农夫约翰被通知,他的一只奶牛逃逸了!所以他决定,马上幽发,尽快把那只奶牛抓回来,他们都站在数轴上.约翰在N(O≤N≤100000)处,奶牛在K(O≤K≤100000)处.约翰有两种办法移动,步行和瞬移:步行每秒种可以让约翰从x处走到x+1或x-1处;而瞬移则可让他在1秒内从x处消失,在2x处出现.然而那只逃逸的奶牛,悲剧地没有发现自己的处境多么糟糕,正站在那儿一动不动.那么,约翰需要多少时间抓住那只牛呢?
    Input

    Line 1: Two space-separated integers: N and K
    仅有两个整数N和K.
    Output
    Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
    最短的时间.
    Sample Input
    5 17
    Farmer John starts at point 5 and the fugitive cow is at point 17.
    Sample Output
    4
    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.
    思路
    这道题依然是一个BFS,可以拓展三种状态:-1,+1或*2,但是要添加一些判断使得不会让标记数组越界导致RE(我死了很多次!!!)
    其他的就是模板了…

    代码

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 int n,k;
     4 bool vis[10000010];
     5 struct node
     6 {
     7     int x,s;
     8 };
     9 void bfs()
    10 {
    11     queue<node> q;
    12     vis[n]=1;
    13     int tx;
    14     q.push((node){n,0});
    15     while(!q.empty())
    16     {
    17         node now=q.front();
    18         //q.pop();
    19         if(now.x==k)
    20         {
    21             cout<<now.s<<endl;
    22             exit(0);
    23         }
    24         tx=now.x-1;
    25         if(tx>=0)
    26         if(!vis[tx])
    27         {
    28             q.push((node){tx,now.s+1});
    29         }
    30         if(tx>2*k)continue;
    31         tx=now.x+1;
    32         if(!vis[tx])
    33         {
    34             q.push((node){tx,now.s+1});
    35             vis[tx]=1;
    36         }
    37         tx=now.x*2;
    38         if(!vis[tx])
    39         {
    40             q.push((node){tx,now.s+1});
    41             vis[tx]=1;
    42         }
    43     }
    44 }
    45 int main()
    46 {
    47     cin>>n>>k;
    48     bfs();
    49     cout<<"这组数据无解"<<endl;//打着玩的
    50     return 0;
    51 }
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  • 原文地址:https://www.cnblogs.com/moyujiang/p/11213630.html
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