Openjudge 2971 grab the bull.

Source: Internet
Author: User

Total time limit:
Memory Limit:

The farmer knows the position of a bull and wants to catch it. Both the farmer and the ox are located on the axis, the farmer starts at point N (0<=n<=100000) and the ox is at point K (0<=k<=100000). Farmers have two ways of moving:

1. Move from X to X-1 or x+1, one minute per move 2, moving from X to 2*x, each move takes a minute Suppose the cow is not aware of the farmer's actions and stands still. How long does it take at least for the farmer to catch the bull?

two integers, N and K
an integer, the minimum number of minutes a farmer can take to catch a cow.
Sample input
5 17
Sample output

The ordinary BFS. Be careful to prune and open large arrays.

1#include <iostream>2#include <algorithm>3#include <cstdio>4#include <cmath>5 using namespacestd;6 intn,k;7 BOOLvis[300001];8 intpos[1000000];9 intstep[1000000];Ten voidBFS () { One     intHd=0, tl=1; Apos[++hd]=N; -step[hd]=0; -      while(hd<=TL) { the         intnow=POS[HD]; -         if(now==k) { -printf"%d\n", STEP[HD]); -             return; +         } -         intnext=now*2; +         if(next>=0&& next<=100000&&!Vis[next]) { Avis[next]=1; atpos[++tl]=Next; -step[tl]=step[hd]+1;} -next=now+1; -         if(next<=100000&&!Vis[next]) { -pos[++tl]=Next; -vis[next]=1; instep[tl]=step[hd]+1;} -next=now-1; to         if(next>=0&&!Vis[next]) { +vis[next]=1; -pos[++tl]=Next; thestep[tl]=step[hd]+1;} *hd++; $     }Panax Notoginseng } - intMain () { thescanf"%d%d",&n,&k); + BFS (); A     return 0; the}

Openjudge 2971 grab the bull.

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