Catch That Cow
Time Limit: 2000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 327864-bit integer IO format: %lld Java class name: Main
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
Source
解题：搜。。。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib> 10 #include <string> 11 #include <set> 12 #include <stack> 13 #define LL long long 14 #define pii pair<int,int> 15 #define INF 0x3f3f3f3f 16 using namespace std; 17 struct node{ 18 int p,step; 19 node(int x = 0,int y = 0):p(x),step(y){} 20 }; 21 int n,k; 22 bool vis[100100]; 23 queue<node>q; 24 int bfs(){ 25 while(!q.empty()) q.pop(); 26 memset(vis,false,sizeof(vis)); 27 vis[n] = true; 28 q.push(node(n,0)); 29 int tmp; 30 while(!q.empty()){ 31 node now = q.front(); 32 q.pop(); 33 if(now.p == k) return now.step; 34 tmp = now.p+1; 35 if(tmp <= 100000 && !vis[tmp]){ 36 vis[tmp] = true; 37 q.push(node(tmp,now.step+1)); 38 } 39 tmp = now.p-1; 40 if(tmp >= 0 && !vis[tmp]){ 41 vis[tmp] = true; 42 q.push(node(tmp,now.step+1)); 43 } 44 tmp = now.p<<1; 45 if(tmp <= 100000 && !vis[tmp]){ 46 vis[tmp] = true; 47 q.push(node(tmp,now.step+1)); 48 } 49 } 50 return -1; 51 } 52 int main() { 53 while(~scanf("%d %d",&n,&k)) 54 printf("%d ",bfs()); 55 return 0; 56 }