CSU 1552 Friends（二分图 + 米勒测试）

Posted 2020-09-12 Dh_q

题目链接：http://acm.csu.edu.cn/csuoj/problemset/problem?pid=1552

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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

Hint

Source

 

题意：

　　给你n个数，两个数相加为素数的时候，就可以成为朋友，选过的数字不能重复选择。

题解：

　　2分图最大匹配问题，和米勒测试。

 

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cstdlib>
  5 #include <string>
  6 #include <vector>
  7 #include <map>
  8 #include <set>
  9 #include <queue>
 10 #include <sstream>
 11 #include <algorithm>
 12 using namespace std;
 13 #define pb push_back
 14 #define mp make_pair
 15 #define ms(a, b)  memset((a), (b), sizeof(a))
 16 //#define LOCAL
 17 #define eps 0.0000001
 18 #define LNF (1<<60)
 19 typedef long long LL;
 20 const int inf = 0x3f3f3f3f;
 21 const int maxn = 100+10;
 22 const int mod = 1e9+7;
 23 LL a[maxn];
 24 bool Map[maxn][maxn], vis[maxn];
 25 int lin[maxn];
 26 LL big_rand(LL m)
 27 {
 28     LL x = rand();
 29     x*=rand();
 30     if(x<0) x-=x;
 31     return x%=m;
 32 }
 33 LL mod_mul(LL x, LL y, LL n)
 34 {
 35     if(x == 0 || y == 0)    return 0;
 36     return (((x&1)*y)%n+(mod_mul(x>>1, y, n)<<1)%n)%n;
 37 }
 38 LL mod_exp(LL x, LL y, LL n)
 39 {
 40     LL ret = 1;
 41     while(y){
 42         if(y&1) ret = mod_mul(ret, x, n);
 43         x = mod_mul(x, x, n);
 44         y >>= 1;
 45     }
 46     return ret;
 47 }
 48 bool Miller_Rabbin(LL n)
 49 {
 50     LL i, j, x, m, k;
 51     if(n==2)    return true;
 52     if(n<2|| !(n&1))    return false;
 53     m = n - 1;k = 0;
 54     while(!(m&1))   m >>= 1, k++;
 55     for(i=0;i<4;i++){
 56         x = big_rand(n-2) + 2;
 57         x = mod_exp(x, m, n);
 58         if(x == 1)  continue;
 59         for(j = 0;j<k;j++){
 60             if(x==n-1)  break;
 61             x = mod_mul(x, x, n);
 62         }
 63         if(j>=k)    return false;
 64     }
 65     return true;
 66 }
 67 bool dfs(int x, int n){
 68     for(int j = 1;j<=n;j++){
 69         if(Map[x][j]&&!vis[j]){
 70             vis[j] = 1;
 71             if(lin[j]==0 || dfs(lin[j], n)){
 72                 lin[j] = x;
 73                 return 1;
 74             }
 75         }
 76     }
 77     return 0;
 78 }
 79 int main()
 80 {
 81 #ifdef LOCAL
 82     freopen("input.txt", "r", stdin);
 83 //      freopen("output.txt", "w", stdout);
 84 #endif // LOCAL
 85 
 86     int n;
 87     while(~scanf("%d", &n)){
 88         ms(Map, 0);
 89         for(int i=1;i<=n;i++)    scanf("%lld", &a[i]);
 90         for(int i=1;i+1<=n;i++){
 91             for(int j=i+1;j<=n;j++){
 92                 if(Miller_Rabbin(a[i]+a[j])){
 93                     Map[i][j] = Map[j][i] = 1;
 94                 }
 95             }
 96         }
 97         int ans = 0;
 98         ms(lin, 0);
 99         for(int i=1;i<=n;i++){
100             ms(vis, 0);
101             if(dfs(i, n))  ans++;
102         }
103         printf("%d\n", ans/2);
104     }
105     return 0;
106 }

View Code

 

将出2分图讲解，和米勒测试。未完待续。。XD