CSU 1552: Friends 图论匹配+超级大素数判定

1552: Friends

Time Limit: 3 Sec  Memory Limit: 256 MB
Submit: 163  Solved: 34
[Submit][Status][Web Board]

Description

On
an alien planet, every extraterrestrial is born with a number. If the
sum of two numbers is a prime number, then two extraterrestrials can be
friends. But every extraterrestrial can only has at most one friend. You
are given all number of the extraterrestrials, please determining the
maximum number of friend pair.

Input

There are several test cases.
Each test start with positive integers N(1 ≤ N ≤ 100), which means there are N extraterrestrials on the alien planet.
The following N lines, each line contains a positive integer pi ( 2 ≤ pi
≤10^18),indicate the i-th extraterrestrial is born with pi number.
The input will finish with the end of file.

Output

For each the case, your program will output maximum number of friend pair.

Sample Input

3
2
2
3

4
2
5
3
8

Sample Output

1
2

 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <string>
 #include <vector>
 #include <set>
 #include <map>
 #include <stack>
 #include <queue>
 #include <sstream>
 #include <iomanip>
 #include <ctime>
 using namespace std;
 typedef long long LL;
 const int INF=0x4fffffff;
 const int EXP=1e-;
 const int MS=;

 int vis[MS];
 int edge[MS][MS];

 int lin[MS];
 LL a[MS];
 int n;

 //
 LL MIN;
 LL mult_mod(LL a,LL b,LL n)
 {
     LL s=;
     while(b)
     {
         if(b&) s=(s+a)%n;
         a=(a+a)%n;
         b>>=;
     }
     return s;
 }

 LL pow_mod(LL a,LL b,LL n)
 {
     LL s=;
     while(b)
     {
         if(b&) s=mult_mod(s,a,n);
         a=mult_mod(a,a,n);
         b>>=;
     }
     return s;
 }

 bool Prime(LL n)
 {
     LL u=n-,pre,x;
     int i,j,k=;
     if(n==||n==||n==||n==||n==)  return ;
     if(n==||(!(n%))||(!(n%))||(!(n%))||(!(n%))||(!(n%)))   return ;
     for(;!(u&);k++,u>>=);
     srand((LL)time());
     for(i=;i<;i++)
     {
         x=rand()%(n-)+;
         x=pow_mod(x,u,n);
         pre=x;
         for(j=;j<k;j++)
         {
             x=mult_mod(x,x,n);
             if(x==&&pre!=&&pre!=(n-))
                 return ;
             pre=x;
         }
         if(x!=)  return false;
     }
     return true;
 }

 bool  dfs(int x)     //   dfs增广路
 {
       for(int i=;i<=n;i++)
       {
             if(edge[x][i]&&!vis[i])
             {
                   vis[i]=;
                   if(lin[i]==-||  dfs(lin[i]))     // i没有匹配 或匹配了可以得到增广路
                   {
                         lin[i]=x;
                         return true;
                   }
             }
       }
       return false;
 }

 int main()
 {
       while(scanf("%d",&n)!=EOF)
       {
             memset(edge,,sizeof(edge));
             for(int i=;i<=n;i++)
                   scanf("%lld",&a[i]);
             for(int i=;i<=n;i++)
                   for(int j=i+;j<=n;j++)
                         if(Prime(a[i]+a[j]))
                               edge[i][j]=edge[j][i]=;
             int ans=;
             memset(lin,-,sizeof(lin));
             for(int i=;i<=n;i++)
             {
                   memset(vis,,sizeof(vis));
                   if(dfs(i))
                         ans++;
             }
             printf("%d\n",ans/);
       }
       return ;
 }