Poj 3278 catch that cow

Catch that cowtime limit: 2000 msmemory limit: 65536 kbthis problem will be judged on PKU. Original ID: 3278
64-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 pointN(0 ≤N≤ 100,000) on a number line and the cow is at a pointK(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 pointXTo the pointsX-1 orX+ 1 in a single minute * teleporting: FJ can move from any pointXTo the point 2 ×XIn 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?

Inputline 1: two space-separated integers: NAnd KOutputline 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

Sourceusaco 2007 Open silver solution: search...

 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 long14 #define pii pair<int,int>15 #define INF 0x3f3f3f3f16 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\n",bfs());55     return 0;56 }

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Poj 3278 catch that cow