POJ 3278 Catch That Cow

题目链接：http://poj.org/problem?id=3278

题目大意：约翰在点N，牛在K，约翰抓牛，约翰一秒走到N+1，N-1或N*2，问多长时间抓到牛。

解题思路：因为让我加深了对DP的看法所以写的。由于可以向前，向后以及翻跟头走二倍，因此会出现跳到后面再回到前面的情况，那么这就不符合DP的条件了。但是实际上跳到后面也是从前面跳过去的，因此相当于是从前面多走了一次然后走到这一步。

代码：

const int maxn = 1e5 + 5;
int n, k;
int dp[maxn];

int solve(){
    memset(dp, 0x3f, sizeof(dp));
    for(int i = 0; i <= k; i++) dp[i] = abs(n - i);
    for(int i = n; i <= 100000; i++){
        if(i & 1){
            if(dp[i - 1] + 1< dp[i]) dp[i] = dp[i - 1] + 1;
            if(dp[(i + 1) / 2] + 2< dp[i]) dp[i] = dp[(i + 1) / 2] + 2; 
            if(dp[i / 2] + 2 < dp[i]) dp[i] = dp[i / 2] + 2;
        }
        else{
            if(i > 0 && dp[i - 1] + 1 < dp[i]) dp[i] = dp[i - 1] + 1;
            if(dp[(i + 2) / 2] + 3 < dp[i]) dp[i] = dp[(i + 2) / 2] + 3;
            if(dp[i / 2] + 1 < dp[i]) dp[i] = dp[i / 2] + 1;
        }
    }
    return dp[k];
}

int main(){
    scanf("%d %d", &n, &k);
    if(n >= k) 
        printf("%d
", n - k);
    else{
        int ans = solve();
        cout << ans << endl;
    }
    return 0;
}

题目：

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 96558		Accepted: 30300

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