HDU 2717 Catch That Cow<BFS+界限优化>

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16152    Accepted Submission(s): 4833

Problem 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.
 

 

Source

USACO 2007 Open Silver

 

Recommend

teddy   |   We have carefully selected several similar problems for you:  2102 1372 1240 1072 1180


BFS水题，但是如果一直报RE的话，注意一点，当我们一直遍历*2的情况的时候，可能会超出数组范围，从而导致RE
所以，设置一个新的界限：所有的pos不会大于2000000，因为数据最大才1000000,所以在这种数据下，你顶多乘以一次2，后面再怎么操作都不可能是最优值
其他的地方就是模板吧。

#include <cstdio>#include <cstring>#include <cstdlib>#include<stack>#include<queue>#include<map>#include<iostream>using namespace std;int x,y;struct Node{    int pos,steps;    friend bool operator < (Node a,Node b)    {        return a.steps>b.steps;    }} node;priority_queue<Node> q;int vis[200005];void bfs(){   node.pos=x;   node.steps=0;    while(!q.empty()) q.pop();    q.push(node);    vis[x]=1;    while(!q.empty())    {        Node tp,tmp=q.top();        q.pop();        if(tmp.pos==y)        {            printf("%d\n",tmp.steps);            return ;        }        for(int i=0;i<3;i++){            if(i==0)                tp.pos=tmp.pos+1;            else                if(i==1)                tp.pos=tmp.pos-1;            else                tp.pos=2*tmp.pos;              if(tp.pos>200000)//多一点这个判断,大于200000就不用入栈                continue;            if(vis[tp.pos])                continue;                tp.steps=tmp.steps+1;                vis[tp.pos]=1;                q.push(tp);        }    }}int main(){    while(~scanf("%d %d",&x,&y))    {        for(int i=0;i<200005;i++)            vis[i]=0;        bfs();    }    return 0;}