Prime Path
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.
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
题目类型：素数+BFS
算法分析：首先将四位数的所有素数找出来，然后从开始的素数每次进行一次BFS搜索，转移的条件是下一个素数的与当前素数只有一位不同，注意每次判断素数的范围是1000~9999!!!
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#include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; const int INF = 0x7FFFFFFF; const int MOD = 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 10000 + 66; struct Node { int v, s; Node () {} Node (int vv, int ss) : v (vv), s (ss) {} }; bool prime[maxn+66], v[maxn+66]; long long res; int aa[16], lena, bb[16], lenb; bool Ok (int a, int b) { lena = 0; do { aa[lena++] = a % 10; a /= 10; } while (a); lenb = 0; do { bb[lenb++] = b % 10; b /= 10; } while (b); int cnt = 0; for (int i = 0; i < lena; i++) if (aa[i] != bb[i]) cnt++; if (cnt == 1) return true; return false; } bool bfs (int a, int b) { res = 0; memset (v, false, sizeof (v)); v[a] = true; queue <Node> qu; qu.push(Node (a, 0)); while (!qu.empty ()) { Node tt = qu.front(); qu.pop(); if (tt.v == b) { res = tt.s; return true; } for (int i = 1000; i < 10000; i++) { if (prime[i] && !v[i] && Ok (tt.v, i)) { v[i] = true; qu.push(Node (i, tt.s + 1)); } } } return false; } void Get () { memset (prime, true, sizeof (prime)); for (int i = 2; i <= maxn; i++) { if (prime[i]) { for (int j = 2; j * i <= maxn; j++) prime[j*i] = false; } } } int main() { // ifstream cin ("aaa.txt"); Get (); int t; cin >> t; while (t--) { int a, b; cin >> a >> b; if(a > b) swap (a, b); if (bfs (a, b)) cout << res << endl; else cout << "Impossible" << endl; } return 0; } |
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