luogu P2685 [USACO07OPEN]抓牛Catch That Cow

题目描述

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?

有人通知FJ他的逃跑的奶牛的位置，FJ要立即把它抓回来。他现在位于点N (0 ≤ N ≤ 100,000)，而奶牛位于跟他同一直线上的点K (0 ≤ K ≤ 100,000)。FJ有两种方法去抓住奶牛：走或传。

走：FJ能从点X走到X-1或X+1的位置，用时一分钟。

传：FJ能从点X到2*X的位置，用时一分钟。

如果奶牛在原地不动，问FJ至少要多少分钟才能抓住奶牛。

输入输出格式

输入格式：

第一行，两个整数，N和K

输出格式：

FJ抓住奶牛用的最少时间。

输入输出样例

输入样例#1：

5 17

输出样例#1：

4

说明

样例解释：

抓住奶牛最快的路径是5-10-9-18-17，用时4分钟。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstdlib>

using namespace std;
const int N=100010;

struct node{
    int x,step;
}now,nxt,top;
int peo,milk;
bool vis[N];
queue<node>q;

inline int read()
{
    int x=0;char c=getchar();
    while(c<'0'||c>'9')c=getchar();
    while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();
    return x;
}

inline void bfs(int start)
{
    now.x=start;
    now.step=0;
    q.push(now);
    while(!q.empty())
    {
        top=q.front();
        q.pop();
        int x=top.x;
        if(x==milk)
        {
            printf("%d
",top.step);
            exit(0);
        }
        if(x+1>=0&&x+1<=100000&&!vis[x+1])
        {
            vis[x+1]=1;
            nxt.x=x+1;
            nxt.step=top.step+1;
            q.push(nxt);
        }
        if(x-1>=0&&x-1<=100000&&!vis[x-1])
        {
            vis[x-1]=1;
            nxt.x=x-1;
            nxt.step=top.step+1;
            q.push(nxt);
        }
        if(x*2>=0&&x*2<=100000&&!vis[x*2])
        {
            vis[x*2]=1;
            nxt.x=x*2;
            nxt.step=top.step+1;
            q.push(nxt);
        }
    }
}

int main()
{
    peo=read();
    milk=read();
    bfs(peo);
    return 0;
}