Catch That Cow

Time Limit: 2000MS	?	Memory Limit: 65536K
Total Submissions: 70652	?	Accepted: 22216

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.

/* *題意：有一個農民和一頭牛，他們在一個數軸上，牛在k位置保持不動，農戶開始時在n位置。 *設農戶當前在M位置，每次移動時有三種選擇： *1.移動到M-1； *2.移動到M+1位置； *3.移動到M*2的位置。 *問最少移動多少次可以移動到牛所在的位置。所以可以用廣搜來搜索這三個狀態，直到搜索到牛所在的位置。 * */#include<cstdio> #include<cstring> #include<queue> #include<algorithm> using namespace std;const int N = 200100; int n, k; struct node {int x, step; }; queue<node> Q; int vis[N];void BFS() { //搜索關于X的三種狀態int X, STEP;while(!Q.empty()){node tmp = Q.front();Q.pop();X = tmp.x;STEP = tmp.step;if(X == k){printf("%d\n",STEP);return ;}if(X >= 1 && !vis[X - 1]) //要保證減1后有意義,所以要X >= 1{node temp;vis[X - 1] = 1;temp.x = X - 1;temp.step = STEP + 1;Q.push(temp);}if(X <= k && !vis[X + 1]){node temp;vis[X + 1] = 1;temp.x = X + 1;temp.step = STEP + 1;Q.push(temp);}if(X <= k && !vis[X * 2]){node temp;vis[X * 2] = 1;temp.x = 2 * X;temp.step = STEP + 1;Q.push(temp);}} }int main() {while(~scanf("%d%d",&n,&k)){while(!Q.empty()) Q.pop();memset(vis,0,sizeof(vis));vis[n] = 1;node t;t.x = n, t.step = 0;Q.push(t);BFS();}return 0; }

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