Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 76520 | Accepted: 24178 |
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.
题目地址：http://poj.org/problem?id=3278
【题意】人从n点出发找在k处不动的牛，在横轴上人只能从n到n-1，n+1，n*2三种步子，每步用时一分钟，求找到牛的最短用时。
【分析】这一道题是一维的，相对来说简单的很。我们在此处用bfs，但是要分开n>k和n<=k这两种情况。
代码一：
1 #include<cstdio> 2 #include<queue> 3 #include<cstring> 4 using namespace std; 5 int n, k; 6 struct Node{ 7 int num, step; 8 }a, b; 9 int vis[100005]; 10 int bfs() 11 { 12 memset(vis, 0, sizeof(vis)); 13 queue<Node> q; 14 a.num = n; 15 a.step = 0; 16 vis[n] = 1; 17 q.push(a); 18 while(1) 19 { 20 a = q.front(); 21 q.pop(); 22 for(int i = 0; i < 3; i++) 23 { 24 if(i == 0) b.num = a.num + 1; 25 if(i == 1) b.num = a.num - 1; 26 if(i == 2) b.num = a.num * 2; 27 if(b.num == k) 28 return a.step + 1; 29 if(b.num < 100005 && b.num >= 0 && !vis[b.num]) 30 { 31 b.step = a.step + 1; 32 vis[b.num] = 1; 33 q.push(b); 34 } 35 } 36 } 37 } 38 int main() 39 { 40 while(scanf("%d %d", &n, &k) != EOF) 41 { 42 if(n >= k) 43 { printf("%d ", n-k); continue; } 44 printf("%d ", bfs()); 45 } 46 return 0; 47 }
代码二：
1 #include<cstdio> 2 #include<string.h> 3 #include<iostream> 4 using namespace std; 5 int N, K; 6 int a[100005], vis[100005], step[100005]; 7 int dir[3][2]={{1,1},{1,-1},{2,0}}; 8 // 后 前 二倍 9 void BFS(int i) 10 { 11 int front, rear, x, k; 12 front = 0; rear = 0; 13 a[rear++] = i; 14 vis[i] = 1; 15 step[i] = 0; 16 while(front < rear) 17 { 18 i = a[front++]; 19 for(k=0; k<3; k++) 20 { 21 x = i * dir[k][0] + dir[k][1]; 22 if(!vis[x] && x >=0 && x <= 100000) 23 { 24 if(x == K) 25 { 26 step[x] = step[i] + 1; 27 return; 28 } 29 else 30 { 31 vis[x] = 1; 32 a[rear++] = x; 33 step[x] = step[i] + 1; 34 } 35 } 36 } 37 } 38 } 39 int main() 40 { 41 cin >> N >> K; 42 memset(vis, 0, sizeof(vis)); 43 memset(step, 0, sizeof(step)); 44 memset(a, 0, sizeof(a)); 45 BFS(N); 46 cout << step[K] << endl; 47 return 0; 48 }