【lｌｉｇｈｔｏｊ】1141-Number Transformation BFS+查找质因子

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       In this problem, you are given an integer number s. You can transform any integer number A to another integer number B by adding x to A. This x is an integer number which is a prime factor of A (please note that 1 and A are not being considered as a factor of A). Now, your task is to find the minimum number of transformations required to transform s to another integer number t.


Input
Input starts with an integer T (≤ 500), denoting the number of test cases.


Each case contains two integers: s (1 ≤ s ≤ 100) and t (1 ≤ t ≤ 1000).


Output
For each case, print the case number and the minimum number of transformations needed. If it's impossible, then print -1.


Sample Input
2
6 12
6 13
Sample Output
Case 1: 2
Case 2: -1

题意：给出两个数S,T,找出S的质因子，s通过加自己的质因子得到一个数，这个数也加自己的质因子，依次循环，看是否能够等于T.,若能得到，就输出相加次数最少的，若不能输出-1.

AC代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
//预先统计出1~1e3每个数的质因子，如果 s == t,自然是一，
//如果s 和 t 其中一个是 质数者必定为 -1，其他 BFS
int T,t,s,ok,cut;
const int MAX=1e3+10;
const int INF = 0x3f3f3f3f / 2;
int p[MAX],vis[MAX];
vector <int> a[MAX];//用vector来存储每个数的质因子 
struct node
{
int x,nl;//nl表示要加的次数 
};
void prime_factor()//查找质因子 
{
memset(p,0,sizeof(p));
p[1]=1;
for(int i=2;i<MAX;i++)
{
for(int j=i+i;j<MAX;j+=i)
p[j]=1;//用打表法把不是素数的数标记，减少了复杂度 
}
for(int i=2;i<MAX;i++)
{
for(int j=2;j<i;j++)
{
if(i%j==0&&!p[j])
a[i].push_back(j); //把每一个数的质因子存起来 
}
}
}
void bfs()
{
memset(vis,0,sizeof(vis));
queue <node> q;
vis[s]=1;
node o;
o.x =s,o.nl=0;
q.push(o); 
while(!q.empty())
{
o=q.front() ;
q.pop() ;
if(o.x==t){
ok=1;
cut=min(cut,o.nl );//min()是一个函数包含在algorithm头文件里 
continue;
}
for(int i=0;i<a[o.x].size() ;i++)//a[o.x].size()表示第o.x行存了多少个数 
{
node w;
w.x =o.x +a[o.x ][i],w.nl =o.nl+1;
if(w.x <=t&&!vis[w.x]) vis[w.x ]=1,q.push(w); 
}
} 
}
int main()
{
prime_factor();
scanf("%d",&T);
int k=1;
while(T--)
{
scanf("%d%d",&s,&t);
if(s==t) printf("Case %d: 0\n",k++);
else if(!p[s]) printf("Case %d: -1\n",k++);
else 
{
ok=0;
cut=INF;
  bfs();
  if(ok) printf("Case %d: %d\n",k++,cut);
  else printf("Case %d: -1\n",k++);
   }
}
return 0;
}