hdu 1973 Prime Path 解题报告
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Prime Path
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 301 Accepted Submission(s): 186
Problem Description
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!
—I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— 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.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.
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.
— Hmm, in that case I need a computer program to minimize the cost. You don’t know some very cheap software gurus, do you?
—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.
— 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!
—I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— 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.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.
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.
— Hmm, in that case I need a computer program to minimize the cost. You don’t know some very cheap software gurus, do you?
—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
31033 81791373 80171033 1033
Sample Output
670解题代码#include <iostream>#include <stdio.h>#include <cstring>#include <queue>using namespace std;int m,n,flag;int a[10001];int vis[10001];int d[4]= {1,10,100,1000};struct node{ int x; int step;//记录当前转换的步数};void prime()//素数筛{ memset(a,0,sizeof(a)); a[0]=a[1]=1; for(int i=2; i<=101; i++) { if(!a[i]) for(int j=i*i; j<10001; j+=i) a[j]=1; }}queue<node> q;int bfs(int s){ while(!q.empty())// 多组测试用例,可能队列不为空,所以首先要清空队列 q.pop(); node st,dt; st.x=s; st.step=0; vis[st.x]=1; q.push(st); flag=0; while(!q.empty()) { st=q.front(); q.pop(); if(st.x==n) { flag=1; return st.step; } for(int i=0; i<4; i++) for(int j=0; j<=9; j++) { if(i==3 && j==0) continue; else { if(j!=st.x/d[i]%10)//判断该位是不是和现有数字相同 { int tx=((st.x/d[i]/10)*10+j)*d[i]+st.x%d[i];// 通项 if(!a[tx] && !vis[tx]) { vis[tx]=1; dt.x=tx; dt.step=st.step+1; q.push(dt); } } } } } return -1;}int main(){ int cs,count; scanf("%d",&cs); prime(); while(cs--) { memset(vis,0,sizeof(vis)); scanf("%d%d",&m,&n); if(m==n) printf("0\n"); else { count=bfs(m); if(flag) printf("%d\n",count); else printf("Impossible\n");//注意找不着就输出Impossible } } return 0;}
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