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 1033Sample Output
6
7
0解题思路:
利用广度优先算法,枚举每一个素数,得到目标素数的层数就是最短路径
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <string>
#include <queue>
using namespace std;
const int MAXN = 10010; //最大数
int N;
bool tis_Prime[MAXN]; //prime质因子
bool isVisi[MAXN];
void euraSelect() {
tis_Prime[1] = false; //第一个数排除
for (int i = 2; i*i < MAXN; ++i) {
if (tis_Prime[i] == true) {
for (int j = i * i; j < MAXN; j += i) {
tis_Prime[j] = false; //排除
}
}
}
}
int main() {
//打表
memset(tis_Prime, true, sizeof(tis_Prime));
euraSelect();
cin >> N;
string X, Y;
while (N--) {
int minCost = -1;
cin >> X >> Y;
memset(isVisi, false, sizeof(isVisi));
queue<string> bfs_queue;
queue<int> layer_queue;
bfs_queue.push(X);
layer_queue.push(0);
while (!bfs_queue.empty()) {
string curPrime = bfs_queue.front();
int curLayer = layer_queue.front();
bfs_queue.pop(), layer_queue.pop();
if (curPrime == Y) {
minCost = curLayer;
break;
}
for (int i = 9; i >= 1; --i) { //千位
string TempStr = curPrime;
if(TempStr[0] == i+'0') continue;
TempStr[0] = i + '0';
if (tis_Prime[atoi(TempStr.c_str())] && !isVisi[atoi(TempStr.c_str())]) {
isVisi[atoi(TempStr.c_str())] = true;
bfs_queue.push(TempStr);
layer_queue.push(curLayer + 1);
}
}
for (int i = 0; i <= 9; ++i) { //百位
string TempStr = curPrime;
if (TempStr[1] == i + '0') continue;
TempStr[1] = i + '0';
if (tis_Prime[atoi(TempStr.c_str())] && !isVisi[atoi(TempStr.c_str())]) {
isVisi[atoi(TempStr.c_str())] = true;
bfs_queue.push(TempStr);
layer_queue.push(curLayer + 1);
}
}for (int i = 0; i <= 9; ++i) {
string TempStr = curPrime;
if (TempStr[2] == i + '0') continue;
TempStr[2] = i + '0';
if (tis_Prime[atoi(TempStr.c_str())] && !isVisi[atoi(TempStr.c_str())]) {
isVisi[atoi(TempStr.c_str())] = true;
bfs_queue.push(TempStr);
layer_queue.push(curLayer + 1);
}
}for (int i = 1; i <= 9; i+=2) {
string TempStr = curPrime;
if (TempStr[3] == i + '0') continue;
TempStr[3] = i + '0';
if (tis_Prime[atoi(TempStr.c_str())] && !isVisi[atoi(TempStr.c_str())]) {
isVisi[atoi(TempStr.c_str())] = true;
bfs_queue.push(TempStr);
layer_queue.push(curLayer + 1);
}
}
}
if (minCost == -1) {
cout << "Impossible." << endl;
}
else {
cout << minCost << endl;
}
}
system("PAUSE");
return 0;
}
