ACdream 1025 bfs

Transform

Time Limit: 4000/2000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

Submit Statistic Next Problem

Problem Description

One can transform xinto x+d  if dis divisor of x. Find out the minimum number of steps to transform a into b.

Input

Two integers a and b.

(1≤a,b≤105)

Output

The only integer denotes the minimum number of steps. Print −1if impossible.

Sample Input

1 6

Sample Output

3

Hint

1→2→4→6or 1→2→3→6

Source

ftiasch

Manager

nanae

 

题意：x变为 x+d d为x的因子  问由a变为b最少要多少步骤  无法到达 输出-1

 

题解： bfs处理 注意求因子的处理

         if(gg.x<=b&&vis[gg.x]==0)  注意两个比较式的先后顺序   vis数组 re多次 

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<stack>
#include<cmath>
#define ll long long 
#define pi acos(-1.0)
#define mod 1000000007
using namespace std;
int a,b;
struct node
{
    int x;
    int step;
}exm,gg;
queue<node> q;
int vis[100005];
void bfs(int st)
{
    memset(vis,0,sizeof(vis));
    while(!q.empty())
     q.pop();
   exm.x=st;
   exm.step=0;
   vis[st]=1;
   q.push(exm);
   while(!q.empty())
   {
       exm=q.front();
       q.pop();
       if(exm.x==b)
       {
           cout<<exm.step<<endl;
           break;
    }
       for(int i=1;i<=sqrt(exm.x);i++)
       {
           if(exm.x%i==0)
           {
               gg.x=exm.x+i;
               gg.step=exm.step+1;
               if(gg.x<=b&&vis[gg.x]==0)
               {
               q.push(gg);
               vis[gg.x]=1;
            }
            gg.x=exm.x+exm.x/i;
               gg.step=exm.step+1;
               if(gg.x<=b&&vis[gg.x]==0)
               {
               q.push(gg);
               vis[gg.x]=1;
            }
        }
    }
   }  
}
int main()
{
    while(scanf("%d %d",&a,&b)!=EOF)
    {
        if(a<=b)
        bfs(a);
        else
        cout<<"-1"<<endl;
    } 
    return 0;
}