Prime Path
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 25592 | Accepted: 14097 |
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.
— 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.
1033The 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.
1733
3733
3739
3779
8779
8179
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
Source
就是给你两个4位数a,b~让你找到通过变化(只有一位不同的素数)的最短路径来变化成b;
#include<cstdio> #include<cstring> #include<ctime> #include<iostream> #include<algorithm> #include<string> #include<queue> #define LL long long using namespace std; int ss[10005] = { 0 }; int cnt = 0;//素数个数 int zs[10000];//保存着四位素数 void sss()//素数晒先筛出四位的素数 { for (int s = 2; s < 10002; s++) { if (ss[s] == 0) { if (s >= 1000) { zs[cnt++] = s; } for (int k = 2 * s; k < 10002; k += s) { ss[k] = 1; } } } } bool vis[10005] = { 0 };//防止多次便利 int maxn = 99999; int st, en; bool check(int a, int b) { int spot = 0; while (a) { if (a % 10 != b % 10) { spot++; } a = a / 10; b = b / 10; } if (spot == 1) { return 1; } return 0; } struct fuck { int data;//素数的值 int step;//步数 }; queue<fuck>q; void bfs() { while (!q.empty()) { fuck t = q.front(); q.pop(); if (t.data == en) { maxn = t.step; return;//注意这里~~第一次找到之后就立刻return就好(因为这是bfs~最先找到一定路径最短) } for (int s = 0; s < cnt; s++) { if (check(zs[s], t.data) && !vis[zs[s]]) { fuck w; w.data = zs[s]; w.step = t.step + 1; vis[zs[s]] = 1; q.push(w); } } } } int main() { int te; scanf("%d", &te); sss(); while (te--) { while (!q.empty()) { q.pop(); } maxn = 9999; memset(vis, 0, sizeof(vis)); scanf("%d%d", &st, &en); vis[st] = 1; fuck w; w.step = 0; while (1) { break; } w.data = st;
q.push(w); bfs(); if (maxn != 99999) { printf("%d\n", maxn); } else { printf("Impossible\n"); } } return 0; }