POJ 3126 Prime Path 筛法+最短路

网上的题解都是用BFS求最短路的，我这里用了dijkstra，感觉功能太强了点。

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15553		Accepted: 8768

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

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
typedef long long ll;
const int INF =0x3f3f3f3f;
const int maxn= 10000 ;

bool vis[maxn+4];
int dp[maxn+4];

struct Edge
{
int from,to;
Edge(){}
Edge(int from,int to):from(from),to(to){}

};
vector<Edge>edges;
vector<int >G[maxn+5];

struct Node
{
int x,dis;
Node(){}
Node(int x,int dis):x(x),dis(dis){}
bool operator<(const Node y)const
{
return dis>y.dis;
}
};

bool done[maxn+4];
void pre()
{
for(int i=2;i*i<=maxn;i++) if(!vis[i])
{
for(int j=i*i;j<=maxn;j+=i)
{
vis[j]=1;
}
}
}
int st,ed;

bool in(int x)
{
return 1000<=x&&x<=9999;
}

inline void add_edge(int x,int y)
{
edges.push_back(Edge(x,y) );
int m=edges.size();
G[x].push_back(m-1);
}

void dijkstra()
{
memset(dp,0x3f,sizeof dp);
dp[st]=0;
memset(done,0,sizeof done);

priority_queue<Node>q;
q.push(Node(st,0));
while(!q.empty())
{
Node now=q.top();q.pop();
int x=now.x;
if(done[x]) continue;
done[x]=1;
if(x==ed) return;//答案已出，可以结束了
int dis=now.dis;
for(int i=0;i<G[x].size();i++)
{
Edge e =edges[G[x][i]];//我晕，写掉了
int y=e.to;
if(done[y]) continue;
if(dis+1<dp[y])
{
dp[y]=dis+1;
q.push(Node(y ,dp[y]) );
}
}
}

}
void work()
{
for(int i=1000;i<=9999;i++)
{
G[i].clear();
}
edges.clear();

for(int x=1000;x<=9999;x++)
{
int up=x/1000;
for(int i=0;i<=9;i++)
{
int y=x-up*1000+i*1000;
if(!in(y)||vis[y]||x==y) continue;
add_edge(x,y);
}

up=x%1000/100;
for(int i=0;i<=9;i++)
{
int y=x-up*100+i*100 ;
if(!in(y)||vis[y]||x==y) continue;
add_edge(x,y);
}

up=x%100/10;
for(int i=0;i<=9;i++)
{
int y=x-up*10+i*10;
if(!in(y)||vis[y]||x==y) continue;
add_edge(x,y);
}

up=x%10;
for(int i=0;i<=9;i++)
{
int y=x-up+i;
if(!in(y)||vis[y]||x==y) continue;
add_edge(x,y);
}
}
dijkstra();

}
int main()
{
pre();
int T;scanf("%d",&T);
while(T--)
{
scanf("%d%d",&st,&ed);
work();
printf("%d\n",dp[ed]);
}

return 0;
}