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A广搜
<span style="color:#330099;">/* A - 广搜 基础 Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status 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 Hint The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes. By Grant Yuan 2014.7.13 */ #include<iostream> #include<stdio.h> #include<string.h> #include<queue> #include<cstdio> using namespace std; bool a[100003]; typedef struct{ int num; int sum; }dd; queue<dd>q; int n,k; int res; void slove() {int m,count; int x; dd init; while(!q.empty()){ if(q.front().num==n) { res=q.front().sum; break; } m=q.front().num; count=q.front().sum; x=m-1; if(a[x]==0) { init.num=x; init.sum=count+1; q.push(init); a[x]=1; } x=m+1; if(a[x]==0) { init.num=x; init.sum=count+1; q.push(init); a[x]=1; } if(m%2==0){ x=m/2; if(a[x]==0) { init.num=x; init.sum=count+1; q.push(init); a[x]=1; }} q.pop();} } int main() { scanf("%d%d",&n,&k); memset(a,0,sizeof(a)); dd init; init.num=k; init.sum=0; a[k]=1; q.push(init); slove(); cout<<res<<endl; return 0; } </span>
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