BZOJ 1646: [Usaco2007 Open]Catch That Cow 抓住那只牛 (BFS)

Posted by 111qqz on Sunday, April 10, 2016

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1646: [Usaco2007 Open]Catch That Cow 抓住那只牛

Time Limit: 5 Sec  Memory Limit: 64 MB Submit: 915  Solved: 441 [Submit][Status][Discuss]

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z处走到x+l或x-l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

OUTPUT DETAILS:

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.

大概是写得第一道bfs了。

/* ***********************************************
Author :111qqz
Created Time :2016年04月10日 星期日 21时00分01秒
File Name :code/bzoj/1646.cpp
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=1E5+7;
int n,k;
bool vis[N];
int d[N];
void bfs()
{
    ms(d,-1);
    queue<int>q;
    q.push(n);
    vis[n] = true;
    d[n] = 0 ;

    while (!q.empty())
    {
    int px = q.front () ; q.pop();
    if (px==k)
    {
        return ;
    }
    int nxt[3];
    nxt[0] = px-1;
    nxt[1] = px+1;
    nxt[2] = 2*px;
    for ( int i = 0 ; i < 3 ; i++)
    {
        if (nxt[i]>=0&&nxt[i]<=100000&&!vis[nxt[i]])
        {
        q.push(nxt[i]);
        vis[nxt[i]] = true;
        d[nxt[i]] = d[px] + 1;
        }
    }
    }
}
int main()
{
    #ifndef  ONLINE_JUDGE 
    freopen("code/in.txt","r",stdin);
  #endif
	
    scanf("%d %d",&n,&k);
    bfs();
    printf("%d\n",d[k]);

  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}

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