HDU 5399 Too Simple
2015-08-18 17:06
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Problem Description
Rhason Cheung had a simple problem, and asked Teacher Mai for help. But Teacher Mai thought this problem was too simple, sometimes naive. So she ask you for help.
Teacher Mai has m functions f1,f2,⋯,fm:{1,2,⋯,n}→{1,2,⋯,n}(that
means for all x∈{1,2,⋯,n},f(x)∈{1,2,⋯,n}).
But Rhason only knows some of these functions, and others are unknown.
She wants to know how many different function series f1,f2,⋯,fm there
are that for every i(1≤i≤n),f1(f2(⋯fm(i)))=i.
Two function series f1,f2,⋯,fm and g1,g2,⋯,gm are
considered different if and only if there exist i(1≤i≤m),j(1≤j≤n),fi(j)≠gi(j).
Input
For each test case, the first lines contains two numbers n,m(1≤n,m≤100).
The following are m lines.
In i-th
line, there is one number −1 or n space-separated
numbers.
If there is only one number −1,
the function fi is
unknown. Otherwise the j-th
number in the i-th
line means fi(j).
Output
For each test case print the answer modulo 109+7.
Sample Input
Sample Output
Rhason Cheung had a simple problem, and asked Teacher Mai for help. But Teacher Mai thought this problem was too simple, sometimes naive. So she ask you for help.
Teacher Mai has m functions f1,f2,⋯,fm:{1,2,⋯,n}→{1,2,⋯,n}(that
means for all x∈{1,2,⋯,n},f(x)∈{1,2,⋯,n}).
But Rhason only knows some of these functions, and others are unknown.
She wants to know how many different function series f1,f2,⋯,fm there
are that for every i(1≤i≤n),f1(f2(⋯fm(i)))=i.
Two function series f1,f2,⋯,fm and g1,g2,⋯,gm are
considered different if and only if there exist i(1≤i≤m),j(1≤j≤n),fi(j)≠gi(j).
Input
For each test case, the first lines contains two numbers n,m(1≤n,m≤100).
The following are m lines.
In i-th
line, there is one number −1 or n space-separated
numbers.
If there is only one number −1,
the function fi is
unknown. Otherwise the j-th
number in the i-th
line means fi(j).
Output
For each test case print the answer modulo 109+7.
Sample Input
3 3 1 2 3 -1 3 2 1
Sample Output
1 HintThe order in the function series is determined. What she can do is to assign the values to the unknown functions. 这题其实很简单,对于多于两个-1的情况不管之前那个怎么选,只要最后一个固定就可以保证,所以答案就是(n!)^(m-1) m是-1的个数 不过此题坑比较多,要注意判断#pragma comment(linker, "/STACK:1024000000,1024000000") #include<set> #include<map> #include<queue> #include<stack> #include<cmath> #include<cstdio> #include<vector> #include<string> #include<cstring> #include<algorithm> #include<functional> using namespace std; typedef long long LL; const LL base = 1e9 + 7; const int maxn = 105; LL T, n, m, f[maxn], a[maxn][maxn]; inline void read(int &x) { char ch; while ((ch = getchar())<'0' || ch>'9'); x = ch - '0'; while ((ch = getchar()) >= '0' && ch <= '9') x = x * 10 + ch - '0'; } int main() { //read(T); for (int i = f[0] = 1; i <= 100; i++) f[i] = f[i - 1] * i % base; while (scanf("%I64d%I64d", &n, &m) != EOF) { LL tot = 0, ans = 1; for (int i = 1; i <= m; i++) { scanf("%I64d", &a[i][1]); if (a[i][1] == -1) tot++; else for (int j = 2; j <= n; j++) { scanf("%I64d", &a[i][j]); for (int k = j - 1; k; k--) if (a[i][j] == a[i][k]) ans = 0; } } for (int i = 1; i < tot; i++) ans = ans * f % base; if (tot == 0) { for (int i = 1; i <= n; i++) a[0][i] = i; for (int i = m; i; i--) for (int j = 1; j <= n; j++) a[0][j] = a[i][a[0][j]]; for (int i = 1; i <= n; i++) if (a[0][i] != i) ans = 0; } printf("%I64d\n", ans); } return 0; }
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