hdu 5399 Too Simple 2015多校联合训练赛#9 模拟
Too Simple
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 797 Accepted Submission(s): 268
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).
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).
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
Hint
The order in the function series is determined. What she can do is to assign the values to the unknown functions.
Author
xudyh
Source
2015 Multi-University Training Contest 9
如果有-1，且函数合法，那么就有（N！）^(x-1)，x是-1的个数
否则只需要判断函数映射是否会x= f(f(f....f(x))即可
函数合法的条件是：y = f(x) y包含1到n的每个数字
#include
#include
#include
#include
using namespace std;
int f[101][101];
#define ll long long
int jie[101];
int mod = 1000000007;
int ans[101];
int main(){
int n,m;
jie[0] = 1;
for(int i =1; i <= 100;i++)
jie[i] =1ll*jie[i-1]*i%mod;
while(scanf("%d%d",&n,&m)!=EOF){
int res = 0;
for(int i = 0;i < m; i++){
scanf("%d",&f[i][1]);
if(f[i][1] == -1) {
res++;
continue;
}
for(int j = 2;j <= n; j++)
scanf("%d",&f[i][j]);
}
if(res == 0){
for(int i = 1;i <= n; i++)
ans[i] = i;
for(int i = m-1; i >= 0; i--)
for(int j = 1;j <= n; j++)
ans[j] = f[i][ans[j]];
res = 1;
for(int i = 1;i <= n; i++)
if(ans[i] != i ) res = 0;
if(res)puts("1");
else puts("0");
}
else {
int flag = 1;
for(int i = 0;i < m && flag; i++){
if(f[i][1] == -1) continue;
memset(ans,0,sizeof(ans));
for(int j = 1;j <= n;j++)
ans[f[i][j]] = 1;
for(int j = 1;j <= n;j++)
if(ans[j] == 0)
flag = 0;
}
if(flag == 0) puts("0");
else {
ll t = 1;
for(int i = 1;i < res; i++){
t = t*jie[n]%mod;
}
cout<