hdu 6038 Function （思维题搞一搞）

f函数的限制关系形成多个环，b的排列也形成多个循环。f中每个环要合法赋值，环的大小要能被b的某个环的大小整除，方案为b环的大小

。再把f每个环的赋值方案相乘就是答案。例如f有环4,6,8。b环 2,3。ans= (2 )* (2 + 3)*(2)

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
const ll md = 1e9 + 7;
const int MAXN = 1e5 + 10;
int fa1[MAXN], fa2[MAXN];
int sum1[MAXN], sum2[MAXN];
set<int>st1, st2;

int find1(int x)
{
return fa1[x] == x ? x : (fa1[x] = find1(fa1[x]));
}
void merge1(int x, int y)
{
    x = find1(x);
    y = find1(y);
if (x != y){
        fa1[x] = y;
        sum1[y] += sum1[x];
    }
}

int find2(int x)
{
return fa2[x] == x ? x : (fa2[x] = find2(fa2[x]));
}
void merge2(int x, int y)
{
    x = find2(x);
    y = find2(y);
if (x != y){
        fa2[x] = y;
        sum2[y] += sum2[x];
    }
}
int n, m;
int p[MAXN], q[MAXN];
int val[MAXN];
void init()
{
for (int i = 0; i < n; i++){
        fa1[i] = i;
        sum1[i] = 1;
    }
for (int i = 0; i < m; i++){
        fa2[i] = i;
        sum2[i] = 1;
    }
memset(val, 0, sizeof(val));
    st1.clear();
    st2.clear();
}
int main()
{
int cas = 1;
while (scanf("%d%d", &n, &m) != EOF){
        init();
for (int i = 0; i < n; i++)
scanf("%d", &p[i]);
for (int i = 0; i < m; i++){
scanf("%d", &q[i]);
        }
for (int i = 0; i < n; i++){
            merge1(i, p[i]);
        }
for (int i = 0; i < m; i++){
            merge2(i, q[i]);
        }
for (int i = 0; i < n; i++){
if (find1(i) == i){
                st1.insert(i);
            }
        }
int MAX = 0;
for (int j = 0; j < m; j++){
if (find2(j) == j){
                st2.insert(sum2[j]);
                val[sum2[j]]++;
                MAX = max(MAX, sum2[j]);
            }
        }
        ll ans = 1;
        ll tmp = 0;
set<int>::iterator it, it2;
for (it = st1.begin(); it != st1.end(); it++){
            tmp = 0;
for (it2 = st2.begin(); it2 != st2.end(); it2++){
int big = *it2;
if (sum1[*it] % (*it2) == 0){
                    tmp += (*it2)*val[*it2];
                    tmp %= md;
                }
/*for (int j = 1; j*j <= big; j++){
 if (sum1[*it] % j == 0){
 ans += (ll)j *val[*it];
 ans %= md;
 if (j*j == big)break;
 }
 }*/
            }
            ans = ans * tmp;
            ans %= md;
        }
printf("Case #%d: %lld\n", cas++, ans);
    }
}