POJ 3126

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10320		Accepted: 5897

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.

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

Northwestern Europe 2006

 

水题，对1000 ～ 9999 的素数按照题目要求建边，最后找一条单源最短路即可。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>

using namespace std;

#define maxn 10005

struct node {
        int d,u;
        bool operator < (const node & rhs) const {
                return d > rhs.d;
        }
};
int a,b,len = 0;
bool prime[maxn],done[maxn];
int key[maxn],ele[maxn],d[maxn];
vector<int> G[maxn];


bool judge(int x,int y) {
        int sum = 0;
        while(x) {
                sum += (x % 10 == y % 10);
                x /= 10;
                y /= 10;
        }

        return sum == 3;
}
void build() {
        for(int i = 0; i < len; i++) {
                for(int j = i + 1; j < len; j++) {
                        if(judge(ele[i],ele[j])) {
                                //printf("%d %d
",ele[i],ele[j]);
                                G[i].push_back(j);
                                G[j].push_back(i);
                        }
                }
        }
}
void init() {
        for(int i = 3; i <= maxn - 5; i++) {
                prime[i] = i % 2;
        }

        prime[2] = 1;

        for(int i = 2; i <=  maxn - 5; i++) {
                if(!prime[i]) continue;
                for(int j = i; j * i <= maxn - 5; j++) {
                        prime[j * i] = 0;
                }
        }

        for(int i = 1000; i <= 9999; i++) {
                if(prime[i]) {
                        ele[len] = i;
                        key[i] = len++;
                }
        }

        build();

}

void dijkstra(int s) {
        for(int i = 0; i < len; i++) d[i] = maxn;
        d[s] = 0;
        memset(done,0,sizeof(done));

        node t;
        priority_queue<node> q;
        t.d = 0; t.u = s;
        q.push(t);

        while(!q.empty()) {
                node x = q.top(); q.pop();
                int u = x.u;
                if(done[u]) continue;
                done[u] = 1;
                for(int i = 0; i < G[u].size(); i++) {
                        int v = G[u][i];
                        if(d[v] > d[u] + 1) {
                                d[v] = d[u] + 1;
                                t.d = d[v];
                                t.u = v;
                                q.push(t);
                        }
                }
        }

}
int main()
{
    //freopen("sw.in","r",stdin);
    //freopen("sw.out","w",stdout);

    int t;
    init();
    scanf("%d",&t);

    while(t--) {
            scanf("%d%d",&a,&b);
            dijkstra(key[a]);

            if(d[ key[b] ] == maxn) {
                    printf("Impossible
");

            } else {
                    printf("%d
",d[ key[b] ]);
            }

    }
    return 0;
}