lightoj-1141【bfs】

最新推荐文章于 2019-03-20 09:37:36 发布

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探讨通过加素因子转换两整数的最短步骤，采用广度优先搜索算法实现。

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题目链接：点击打开链接

1141 - Number Transformation

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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

Output for Sample Input

6 12

6 13

Case 1: 2

Case 2: -1

大意：题意：给两个数 s ，t ，用 s 加上它的素因子（不包括 1 和它自身）得到另外一个数，然后重复之前的操作，每次加上当前数的素因子。。。直到得到 t 为止，问最少经过多少次的得到 t ，若得不到输出 -1

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
int a,b;
bool shu[1010]={1,1};
bool vis[1010];
struct node
{
	int val,step;
};
int bfs()
{
	memset(vis,0,sizeof(vis));
	queue<node> Q;
	node now,next;
	now.val=a,now.step=0;
	Q.push(now);
	while(!Q.empty())
	{
		now=Q.front();
		Q.pop();
		if(now.val==b)
			return now.step;
		for(int i=2;i<now.val;i++)
		{
			if(now.val%i||shu[i])	continue;
			if(now.val+i>b||vis[now.val+i])	continue;
			vis[now.val+i]=1;
			next.val=now.val+i;
			next.step=now.step+1;
			Q.push(next);
		}
	}
	return -1;
}
int main()
{
	for(int i=2;i<=1000;i++)
	{
		if(shu[i])	continue;
		for(int j=i*2;j<=1000;j+=i)
			shu[j]=1;
	}
	int t,text=0;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&a,&b);
		printf("Case %d: %d\n",++text,bfs());
	}
	return 0;
}