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[];
 void bfs(int st)
 {
     memset(vis,,sizeof(vis));
     while(!q.empty())
      q.pop();
    exm.x=st;
    exm.step=;
    vis[st]=;
    q.push(exm);
    while(!q.empty())
    {
        exm=q.front();
        q.pop();
        if(exm.x==b)
        {
            cout<<exm.step<<endl;
            break;
     }
        for(int i=;i<=sqrt(exm.x);i++)
        {
            if(exm.x%i==)
            {
                gg.x=exm.x+i;
                gg.step=exm.step+;
                if(gg.x<=b&&vis[gg.x]==)
                {
                q.push(gg);
                vis[gg.x]=;
             }
             gg.x=exm.x+exm.x/i;
                gg.step=exm.step+;
                if(gg.x<=b&&vis[gg.x]==)
                {
                q.push(gg);
                vis[gg.x]=;
             }
         }
     }
    }
 }
 int main()
 {
     while(scanf("%d %d",&a,&b)!=EOF)
     {
         if(a<=b)
         bfs(a);
         else
         cout<<"-1"<<endl;
     }
     return ;
 }