题目：
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
题目大意：
给出两个四位的素数，每次更改其中的一位，要求更改后仍是四位的素数，求从第一个素数到第二个素数的最短路径长度。
解题思路：
把所有的四位素数先存储在数组中用于判断素数，对数字的最高位遍历1到9，百位和十位遍历0到9，个位遍历1到9的奇数，BFS遍历，将每次考虑过的素数做标记，以后不再遍历。
代码：
#include <iostream>
#include <set>
#include <cmath>
#include <queue>using namespace std;
int su[]={1009,1013
,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069
,1087,1091,1093,1097,1103,1109,1117,1123,1129,1151
,1153,1163,1171,1181,1187,1193,1201,1213,1217,1223
,1229,1231,1237,1249,1259,1277,1279,1283,1289,1291
,1297,1301,1303,1307,1319,1321,1327,1361,1367,1373
,1381,1399,1409,1423,1427,1429,1433,1439,1447,1451
,1453,1459,1471,1481,1483,1487,1489,1493,1499,1511
,1523,1531,1543,1549,1553,1559,1567,1571,1579,1583
,1597,1601,1607,1609,1613,1619,1621,1627,1637,1657
,1663,1667,1669,1693,1697,1699,1709,1721,1723,1733
,1741,1747,1753,1759,1777,1783,1787,1789,1801,1811
,1823,1831,1847,1861,1867,1871,1873,1877,1879,1889
,1901,1907,1913,1931,1933,1949,1951,1973,1979,1987
,1993,1997,1999,2003,2011,2017,2027,2029,2039,2053
,2063,2069,2081,2083,2087,2089,2099,2111,2113,2129
,2131,2137,2141,2143,2153,2161,2179,2203,2207,2213
,2221,2237,2239,2243,2251,2267,2269,2273,2281,2287
,2293,2297,2309,2311,2333,2339,2341,2347,2351,2357
,2371,2377,2381,2383,2389,2393,2399,2411,2417,2423
,2437,2441,2447,2459,2467,2473,2477,2503,2521,2531
,2539,2543,2549,2551,2557,2579,2591,2593,2609,2617
,2621,2633,2647,2657,2659,2663,2671,2677,2683,2687
,2689,2693,2699,2707,2711,2713,2719,2729,2731,2741
,2749,2753,2767,2777,2789,2791,2797,2801,2803,2819
,2833,2837,2843,2851,2857,2861,2879,2887,2897,2903
,2909,2917,2927,2939,2953,2957,2963,2969,2971,2999
,3001,3011,3019,3023,3037,3041,3049,3061,3067,3079
,3083,3089,3109,3119,3121,3137,3163,3167,3169,3181
,3187,3191,3203,3209,3217,3221,3229,3251,3253,3257
,3259,3271,3299,3301,3307,3313,3319,3323,3329,3331
,3343,3347,3359,3361,3371,3373,3389,3391,3407,3413
,3433,3449,3457,3461,3463,3467,3469,3491,3499,3511
,3517,3527,3529,3533,3539,3541,3547,3557,3559,3571
,3581,3583,3593,3607,3613,3617,3623,3631,3637,3643
,3659,3671,3673,3677,3691,3697,3701,3709,3719,3727
,3733,3739,3761,3767,3769,3779,3793,3797,3803,3821
,3823,3833,3847,3851,3853,3863,3877,3881,3889,3907
,3911,3917,3919,3923,3929,3931,3943,3947,3967,3989
,4001,4003,4007,4013,4019,4021,4027,4049,4051,4057
,4073,4079,4091,4093,4099,4111,4127,4129,4133,4139
,4153,4157,4159,4177,4201,4211,4217,4219,4229,4231
,4241,4243,4253,4259,4261,4271,4273,4283,4289,4297
,4327,4337,4339,4349,4357,4363,4373,4391,4397,4409
,4421,4423,4441,4447,4451,4457,4463,4481,4483,4493
,4507,4513,4517,4519,4523,4547,4549,4561,4567,4583
,4591,4597,4603,4621,4637,4639,4643,4649,4651,4657
,4663,4673,4679,4691,4703,4721,4723,4729,4733,4751
,4759,4783,4787,4789,4793,4799,4801,4813,4817,4831
,4861,4871,4877,4889,4903,4909,4919,4931,4933,4937
,4943,4951,4957,4967,4969,4973,4987,4993,4999,5003
,5009,5011,5021,5023,5039,5051,5059,5077,5081,5087
,5099,5101,5107,5113,5119,5147,5153,5167,5171,5179
,5189,5197,5209,5227,5231,5233,5237,5261,5273,5279
,5281,5297,5303,5309,5323,5333,5347,5351,5381,5387
,5393,5399,5407,5413,5417,5419,5431,5437,5441,5443
,5449,5471,5477,5479,5483,5501,5503,5507,5519,5521
,5527,5531,5557,5563,5569,5573,5581,5591,5623,5639
,5641,5647,5651,5653,5657,5659,5669,5683,5689,5693
,5701,5711,5717,5737,5741,5743,5749,5779,5783,5791
,5801,5807,5813,5821,5827,5839,5843,5849,5851,5857
,5861,5867,5869,5879,5881,5897,5903,5923,5927,5939
,5953,5981,5987,6007,6011,6029,6037,6043,6047,6053
,6067,6073,6079,6089,6091,6101,6113,6121,6131,6133
,6143,6151,6163,6173,6197,6199,6203,6211,6217,6221
,6229,6247,6257,6263,6269,6271,6277,6287,6299,6301
,6311,6317,6323,6329,6337,6343,6353,6359,6361,6367
,6373,6379,6389,6397,6421,6427,6449,6451,6469,6473
,6481,6491,6521,6529,6547,6551,6553,6563,6569,6571
,6577,6581,6599,6607,6619,6637,6653,6659,6661,6673
,6679,6689,6691,6701,6703,6709,6719,6733,6737,6761
,6763,6779,6781,6791,6793,6803,6823,6827,6829,6833
,6841,6857,6863,6869,6871,6883,6899,6907,6911,6917
,6947,6949,6959,6961,6967,6971,6977,6983,6991,6997
,7001,7013,7019,7027,7039,7043,7057,7069,7079,7103
,7109,7121,7127,7129,7151,7159,7177,7187,7193,7207
,7211,7213,7219,7229,7237,7243,7247,7253,7283,7297
,7307,7309,7321,7331,7333,7349,7351,7369,7393,7411
,7417,7433,7451,7457,7459,7477,7481,7487,7489,7499
,7507,7517,7523,7529,7537,7541,7547,7549,7559,7561
,7573,7577,7583,7589,7591,7603,7607,7621,7639,7643
,7649,7669,7673,7681,7687,7691,7699,7703,7717,7723
,7727,7741,7753,7757,7759,7789,7793,7817,7823,7829
,7841,7853,7867,7873,7877,7879,7883,7901,7907,7919
,7927,7933,7937,7949,7951,7963,7993,8009,8011,8017
,8039,8053,8059,8069,8081,8087,8089,8093,8101,8111
,8117,8123,8147,8161,8167,8171,8179,8191,8209,8219
,8221,8231,8233,8237,8243,8263,8269,8273,8287,8291
,8293,8297,8311,8317,8329,8353,8363,8369,8377,8387
,8389,8419,8423,8429,8431,8443,8447,8461,8467,8501
,8513,8521,8527,8537,8539,8543,8563,8573,8581,8597
,8599,8609,8623,8627,8629,8641,8647,8663,8669,8677
,8681,8689,8693,8699,8707,8713,8719,8731,8737,8741
,8747,8753,8761,8779,8783,8803,8807,8819,8821,8831
,8837,8839,8849,8861,8863,8867,8887,8893,8923,8929
,8933,8941,8951,8963,8969,8971,8999,9001,9007,9011
,9013,9029,9041,9043,9049,9059,9067,9091,9103,9109
,9127,9133,9137,9151,9157,9161,9173,9181,9187,9199
,9203,9209,9221,9227,9239,9241,9257,9277,9281,9283
,9293,9311,9319,9323,9337,9341,9343,9349,9371,9377
,9391,9397,9403,9413,9419,9421,9431,9433,9437,9439
,9461,9463,9467,9473,9479,9491,9497,9511,9521,9533
,9539,9547,9551,9587,9601,9613,9619,9623,9629,9631
,9643,9649,9661,9677,9679,9689,9697,9719,9721,9733
,9739,9743,9749,9767,9769,9781,9787,9791,9803,9811
,9817,9829,9833,9839,9851,9857,9859,9871,9883,9887
,9901,9907,9923,9929,9931,9941,9949,9967,9973};set<int> ss(su,su+sizeof(su)/sizeof(*su));
void inttostr(int a,char* ch){
ch[0]=a/1000+48;
ch[1]=(a/100)%10+48;
ch[2]=(a/10)%10+48;
ch[3]=a%10+48;
ch[4]='\0';
}int strtoint(char* const ch){
int ans=0;
for(int i=0;i<4;i++){
ans=ans+(ch[i]-48)*pow(10,3-i);
}
return ans;
}int main(){
int cas;
cin>>cas;
while(cas--){
bool jsu[10002]={false};
for(int i=0;i<1061;i++)jsu[su[i]]=true;
bool vis[10002]={false};
queue<pair<int,int> > qu;
int beg,tag;
int ans=0;
cin>>beg>>tag;
vis[beg]=true;
qu.push(pair<int,int>(beg,0));
while(!qu.empty()){
pair<int,int> cur=qu.front();
qu.pop();
if(cur.first==tag){
ans=cur.second;
break;
}
char cu[5];
inttostr(cur.first,cu);
for(int i=0;i<4;i++){
for(char j='0';j<='9';j++){
if(i==3&&!j%2||i==0&&j=='0')continue;
if(j!=cu[i]){
char c=cu[i];
cu[i]=j;
int icu=strtoint(cu);
if(!vis[icu]){
vis[icu]=true;
if(jsu[icu]){
qu.push(pair<int,int>(icu,cur.second+1));
}
}
cu[i]=c;
}
}
}
}
cout<<ans<<endl;
}
return 0;
}