Poj 3278 Catch That Cow
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Catch That Cow
Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 58054 Accepted: 18053
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
思路:既然我们要找位置的,所以必然涉及到当前位置和步数的问题,所以可以设一个结构体,里面有数据步数和位置,然后搜索,当遇到m时,记录下最短时间,当在不断收索时,遇到比当前最优的时间还多的应立即跳出,位置的标记也是优化的一种形式。
#include <iostream>#include<cstring>#include<cstdio>#include<queue>using namespace std;int n,m;struct node{ int x,ans;} ;int vis[1000010],mi;int jx[]={-1,1};void bfs(){ queue<node>q; memset(vis,0,sizeof(vis)); node f1,f2; int i,j,a; f1.x=n,f1.ans=0; q.push(f1); mi=0x3f3f3f3f; while(!q.empty()) { f1=q.front(); q.pop(); if(f1.x==m) { if(mi>f1.ans) { mi=f1.ans; continue; } } if(f1.ans>=mi) continue; for(i=0;i<3;i++) { f2=f1; if(i==2) f2.x=f2.x*2;//这里与i==1||i==0时都要改变f2.x的值,而不是设一个新值,因为还要把此数压入栈内 else f2.x=f2.x+jx[i]; if(f2.x>=0&&f2.x<=100000&&vis[f2.x]==0) { vis[f2.x]=1; f2.ans++; q.push(f2); } } }}int main(){ ios::sync_with_stdio(false);//注意写上加速语句防TLE while(cin>>n>>m) { if(n==m) printf("0\n"); else { bfs(); cout<<mi<<endl; } } return 0;}
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