2971:抓住那头牛

总时间限制:

2000ms

内存限制:

65536kB

描述

农夫知道一头牛的位置，想要抓住它。农夫和牛都位于数轴上，农夫起始位于点N(0<=N<=100000)，牛位于点K(0<=K<=100000)。农夫有两种移动方式：

1、从X移动到X-1或X+1，每次移动花费一分钟

2、从X移动到2*X，每次移动花费一分钟

 

假设牛没有意识到农夫的行动，站在原地不动。农夫最少要花多少时间才能抓住牛？

输入

两个整数，N和K

输出

一个整数，农夫抓到牛所要花费的最小分钟数

样例输入

5 17

样例输出

4
深搜 炸：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>

using namespace std;
const int N=100010;

bool vis[N];
int people;
int milk;
int answer=99999999;
int nowans=0;

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;
}

void dfs(int nowpeople)
{
    vis[nowpeople]=1;
    if(nowpeople==milk)
    {
        answer=min(answer,nowans);
        return ;
    }
    if(!vis[nowpeople+1]&&nowpeople+1>0&&nowpeople+1<=milk+1)
    {
        nowans++;
        vis[nowpeople+1]=1;
        dfs(nowpeople+1);
        nowans--;
        vis[nowpeople+1]=0;
    }
    if(!vis[nowpeople-1]&&nowpeople-1>0&&nowpeople-1<=milk+1)
    {
        nowans++;
        vis[nowpeople-1]=1;
        dfs(nowpeople-1);
        nowans--;
        vis[nowpeople-1]=0;
    }
    if(!vis[nowpeople*2]&&nowpeople*2>0&&nowpeople*2<=milk+1)
    {
        nowans++;
        vis[nowpeople*2]=1;
        dfs(nowpeople*2);
        nowans--;
        vis[nowpeople*2]=0;
    }
}

int main()
{
    people=read();
    milk=read();
    dfs(people);
    printf("%d",answer);
    return 0;
} 

广搜：

#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;
}