匈牙利最大匹配+大质数判定 csu1552 Friends

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题意：每个人都有一个数字，如果两个人的数字之和是质数，那么就可以成为朋友，一个人最多只能有一个朋友

思路：二分图裸匈牙利最大匹配和大质数判定的模板题，测试模板专用

#include<map>#include<set>#include<cmath>#include<ctime>#include<stack>#include<queue>#include<cstdio>#include<cctype>#include<string>#include<vector>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#define fuck printf("fuck")#define FIN freopen("input.txt","r",stdin)#define FOUT freopen("output.txt","w+",stdout)using namespace std;typedef long long LL;const int MX = 200 + 5;LL multi(LL a, LL b, LL mod) {    LL ret = 0;    while(b) {        if(b & 1) ret = ret + a;        if(ret >= mod) ret -= mod;        a = (a + a) % mod;        if(a >= mod) a -= mod;        b >>= 1;    }    return ret;}LL power(LL a, LL b, LL mod) {    LL ret = 1;    while(b) {        if(b & 1) ret = multi(ret, a, mod);        a = multi(a, a, mod);        b >>= 1;    }    return ret;}bool Miller_Rabin(LL n) {    LL u = n - 1, pre, x;    int i, j, k = 0;    if(n == 2 || n == 3 || n == 5 || n == 7 || n == 11)  return true;    if(n == 1 || (!(n % 2)) || (!(n % 3)) || (!(n % 5)) || (!(n % 7)) || (!(n % 11)))   return false;    for(; !(u & 1); k++, u >>= 1);    srand(time(NULL));    for(i = 0; i < 5; i++) {        x = rand() % (n - 2) + 2;        x = power(x, u, n);        pre = x;        for(j = 0; j < k; j++) {            x = multi(x, x, n);            if(x == 1 && pre != 1 && pre != (n - 1))                return false;            pre = x;        }        if(x != 1)  return false;    }    return true;}LL A[MX];int match[MX];bool vis[MX], G[MX][MX];bool DFS(int n, int u) {    vis[u] = true;    for(int i = 1; i <= n; i++) {        if(!G[u][i]) continue;        int v = i, w = match[v];        if(w < 0 || !vis[w] && DFS(n, w)) {            match[v] = u;            match[u] = v;            return true;        }    }    return false;}int BM(int n) {    int ret = 0;    memset(match, -1, sizeof(match));    for(int i = 1; i <= n; i++) {        if(match[i] < 0) {            memset(vis, 0, sizeof(vis));            if(DFS(n, i)) ret++;        }    }    return ret;}int main() {    int n; //FIN;    while(~scanf("%d", &n)) {        memset(G, 0, sizeof(G));        for(int i = 1; i <= n; i++) {            scanf("%lld", &A[i]);        }        for(int i = 1; i <= n; i++) {            for(int j = i + 1; j <= n; j++) {                if(Miller_Rabin(A[i] + A[j])) {                    G[i][j] = G[j][i] = 1;                }            }        }        printf("%d\n", BM(n));    }    return 0;}

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