Function
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1594 Accepted Submission(s): 302
Problem Description
You are given a permutation a from 0 to n−1 and
a permutation b from 0 to m−1.
Define that the domain of function f is the set of integers from 0 to n−1, and the range of it is the set of integers from 0 to m−1.
Please calculate the quantity of different functions f satisfying that f(i)=bf(ai) for each i from 0 to n−1.
Two functions are different if and only if there exists at least one integer from 0 to n−1 mapped into different integers in these two functions.
The answer may be too large, so please output it in modulo 109+7.
Define that the domain of function f is the set of integers from 0 to n−1, and the range of it is the set of integers from 0 to m−1.
Please calculate the quantity of different functions f satisfying that f(i)=bf(ai) for each i from 0 to n−1.
Two functions are different if and only if there exists at least one integer from 0 to n−1 mapped into different integers in these two functions.
The answer may be too large, so please output it in modulo 109+7.
Input
The input contains multiple test cases.
For each case:
The first line contains two numbers n, m. (1≤n≤100000,1≤m≤100000)
The second line contains n numbers, ranged from 0 to n−1, the i-th number of which represents ai−1.
The third line contains m numbers, ranged from 0 to m−1, the i-th number of which represents bi−1.
It is guaranteed that ∑n≤106, ∑m≤106.
For each case:
The first line contains two numbers n, m. (1≤n≤100000,1≤m≤100000)
The second line contains n numbers, ranged from 0 to n−1, the i-th number of which represents ai−1.
The third line contains m numbers, ranged from 0 to m−1, the i-th number of which represents bi−1.
It is guaranteed that ∑n≤106, ∑m≤106.
Output
For each test case, output "Case #x: y"
in one line (without quotes), where x indicates
the case number starting from 1 and y denotes
the answer of corresponding case.
Sample Input
3 2
1 0 2
0 1
3 4
2 0 1
0 2 3 1
Sample Output
Case #1: 4
Case #2: 4
題目大意：給你一些個關係式f(i)=bf(a(i)),詢問滿足這個關係式有多少種可能。
解題思路：
其實這道題可以轉化爲求環的問題，有多少種方式可以構成題目要求的環，即b的環可以有多少種方式畫成a的環
來看第一個樣例：
a：1 ， 0 ， 2
b：0 ， 1
則a0->a1->a0, a2->a2
b0->b0, b1->b0
即 ![]()
則，由圖可以得到a的第一環可以由b的兩種方式構成，a的第二個環可以由b的兩種方式構成，則由b構成a總共有2*2=4種方式。
讓我們看第二個樣例：
a：2 ， 0 ， 1
b: 0 ， 2 ， 3 ， 1
則a：a0->a2->a1->a0
b: b0->b0 b1->b2->b3->b1
即：![]()
如圖，a只有一種環，所以b只需要構成這一種環就行了，有4種方式可以構成這個環。
所以題目轉換爲求構成環的方式了。
ac代碼：
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define mkp make_pair
#define ite iterator
#define fi first
#define se second
#define FOR(i, l, r) for(int i = l; i <= r; i++)
#define ROF(i, r, l) for(int i = r; i >= l; i--)
#define M 1000000007
const int maxn = 100100;
int n, m, cal[2][maxn], a[maxn], b[maxn], ans = 1;
bool vis[maxn];
void dfs(int t, int l, int *a, int k){
if(vis[t]){
cal[k][l]++;
return;
}
vis[t] = 1;
dfs(a[t], l + 1, a, k);
}
int main(){
int ca = 0;
while(~scanf("%d%d", &n, &m)){
for(int i = 0; i < n; i++)
scanf("%d", a + i);
for(int i = 0; i < m; i++)
scanf("%d", b + i);
memset(cal, 0, sizeof(cal));
memset(vis, 0, sizeof(vis));
for(int i = 0; i < m; i++)
if (!vis[i]) dfs(i, 0, b, 0);
memset(vis, 0, sizeof(vis));
for(int i = 0; i < n; i++)
if (!vis[i]) dfs(i, 0, a, 1);
ans = 1;
for(int i = 1; i <= n; i++)
if(cal[1][i]){
int lim = (int)sqrt(i + 0.5), ta = 0;
for(int j = 1; j <= lim; j++)
if(i % j == 0){
(ta += (ll)cal[0][j] % M * j % M) %= M;
if (j * j != i) (ta += (ll)cal[0][i / j] % M * (i / j) % M) %= M;
}
for(int j = 1; j <= cal[1][i]; j++)
ans = (ll)ans * ta % M;
}
printf("Case #%d: %d\n", ++ca, ans);
}
return 0;
}