POJ - 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 90830		Accepted: 28508

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

傳送門：http://poj.org/problem?id=3278

題意：農夫所在位置爲N，0≤N≤100000，奶牛所在位置爲K，0≤K≤100000，農夫和奶牛在同一條直線上，農夫每次移動可以向前走一步，或者退後一步，或者傳送到當前座標的2倍位置，比如從5到10，一步完成。奶牛並不知道農夫在抓它，所以一直在原地不動，問農夫最短可以用多少步抓到奶牛。

分析：一道BFS的題，但是DP也可以解，每次BFS都進行三次操作，前進一步，後退一步，傳送到二倍位置，如果沒有訪問過，則標記訪問，步數加一，直到找到解。因爲農夫後退只能一步一步退，所以用DP的時候先從農夫的初始位置向前掃描，每移動一次步數加一直到到0位置。

for(i=start;i>0;i--)
    	dp[i-1]=dp[i]+1;

再從初始位置向後掃描，後位置(dp[i])步數爲前一位置步數+1(dp[i-1]+1)，再計算從二分之一當前座標傳送到目前位置的步數，取最小值，注意細節處理。當i爲奇數時dp[i]=min(dp[i],dp[i/2+1]+2);例如5需要3->6->5，一共兩步操作，當i爲偶數時dp[i]=min(dp[i],dp[i/2]+1);掃描完成後dp[奶牛初始位置]就爲最終答案。

經過測試DP和BFS內存使用不相上下，但DP4ms就能AC，BFS需要32ms。

細節：第一次DP數組只開了100000，一直WA，後來發現是數組不夠大，所以遇到這種類型的還是把數組多開幾個單位比較保險。

/*DP*/
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int MAX = 1e5+5;
int start,target;
int now ,pos;
int dp[MAX];
int i;
int main()
{
    cin >> start >> target;
    for(i=start;i>0;i--)
        dp[i-1]=dp[i]+1;
    for(i=start+1;i<=target;i++)
    {
        dp[i]=dp[i-1]+1;
        if (i%2==1) dp[i]=min(dp[i],dp[i/2+1]+2);
        else
            dp[i]=min(dp[i],dp[i/2]+1);
    }
    cout << dp[target] <<endl;
    return 0;
}

/*BFS*/
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int MAX=1E5+10;
static bool visit[MAX]={false};
static int step[MAX]={0};
int now,position;
int first,finally;
queue <int> que;
int BFS(int n)
{
    que.push(n);
    while(!que.empty())
    {
        now=que.front();
        que.pop();
        for(int i=0;i<3;i++)
        {
            if (i==0)   position=now-1;
            else if(i==1) position=now+1;
            else position=2*now;//now 當前位置 position 經過一步移動後的位置
            if(visit[position]==false&&position>=0&&position<=MAX)
            {
                visit[position]=true;
                step[position]=step[now]+1;
                que.push(position);
            }
            if(position==finally)
                return step[position];
        }
    }
}
int main()
{
    cin >> first >> finally;
    while(!que.empty()) que.pop();
    memset(visit,false,sizeof(visit));
    memset(step,0,sizeof(step));
    if (first>=finally) cout << first-finally <<endl;
    else
        cout << BFS(first) << endl;
    return 0;
}