POJ 3278 Catch That Cow
2011-08-23 02:44
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POJ_3278[/b]
这个是个比较简单的BFS的题目。
这个是个比较简单的BFS的题目。
#include<stdio.h> #include<string.h> int dis[200010],q[200010]; int main() { int i,j,N,K,front,rear,n,newn; while(scanf("%d%d",&N,&K)==2) { memset(dis,-1,sizeof(dis)); dis =0; front=rear=0; q[rear++]=N; while(front<rear) { n=q[front++]; if(n==K) break; if(n<K) { newn=n-1; if(dis[newn]<0) { dis[newn]=dis +1; q[rear++]=newn; } newn=n+1; if(dis[newn]<0) { dis[newn]=dis +1; q[rear++]=newn; } newn=2*n; if(dis[newn]<0) { dis[newn]=dis +1; q[rear++]=newn; } } else { newn=n-1; if(dis[newn]<0) { dis[newn]=dis +1; q[rear++]=newn; } } } printf("%d\n",dis[K]); } return 0; }
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