在a-table的基礎上實現拖拽功能 (封裝的元件)
阿新 • • 發佈:2021-01-21
技術標籤:廣搜
The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices.
— It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— In fact, I do. You see, there is this programming contest going on… Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
1033
1733
3733
3739
3779
8779
8179
The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.
Input
One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).
Output
One line for each case, either with a number stating the minimal cost or containing the word Impossible.
Sample Input
3
1033 8179
1373 8017
1033 1033
Sample Output
6
7
0
題意 把A變成B 變換過程的每個中間值 都必須是素數 因為是四位數 可以將每一位分離開 每一位都可以0-9取值
#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdio.h>
#include<queue>
using namespace std;
const int N=10010;
int book[N],vis[N];
struct node
{
int now,step;
};
void prime()
{
memset(book,0,sizeof(book));
book[1]=1;
for(int i=2;i<N;i++)
{
if(!book[i])
{
for(int j=i*2;j<N;j+=i)
{
book[j]=1;
}
}
}
}
int bfs(int x,int y)
{
int num;
queue<node>Q;
node q,s;
q.now=x;
q.step=0;
Q.push(q);
while(!Q.empty())
{
q=Q.front();
Q.pop();
if(q.now==y)
return q.step;
int b[5];
b[1]=q.now/1000;
b[2]=q.now/100%10;
b[3]=q.now/10%10;
b[4]=q.now%10;
for(int i=1;i<=4;i++)
{
int t=b[i];
for(int j=0;j<=9;j++)
{
if(b[i]!=j)
{
b[i]=j;
num=b[1]*1000+b[2]*100+b[3]*10+b[4];
}
if(num>=1000&&num<=9999&&vis[num]==0&&book[num]==0)
{
vis[num]=1;
s.step=q.step+1;
s.now=num;
Q.push(s);
}
}
b[i]=t;
}
}
return -100;
}
int main()
{
int n;
prime();
scanf("%d",&n);
prime();
while(n--)
{
int a,b;
scanf("%d %d",&a,&b);
memset(vis,0,sizeof(vis));
int t=bfs(a,b);
if(t!=-100)
printf("%d\n",t);
else
printf("Impossible\n");
}
}