# codeforces 341C Iahub and Permutations(结合数dp)

www.MyException.Cn  网友分享于：2013-10-08  浏览：6次
codeforces 341C Iahub and Permutations(组合数dp)
C. Iahub and Permutations
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Iahub is so happy about inventing bubble sort graphs that he's staying all day long at the office and writing permutations. Iahubina is angry that she is no more important for Iahub. When Iahub goes away, Iahubina comes to his office and sabotage his research work.

The girl finds an important permutation for the research. The permutation contains n distinct integers a1a2, ..., an (1 ≤ ai ≤ n). She replaces some of permutation elements with -1 value as a revenge.

When Iahub finds out his important permutation is broken, he tries to recover it. The only thing he remembers about the permutation is it didn't have any fixed point. A fixed point for a permutation is an element ak which has value equal to k (ak = k). Your job is to proof to Iahub that trying to recover it is not a good idea. Output the number of permutations which could be originally Iahub's important permutation, modulo 1000000007 (109 + 7).

Input

The first line contains integer n (2 ≤ n ≤ 2000). On the second line, there are n integers, representing Iahub's important permutation after Iahubina replaces some values with -1.

It's guaranteed that there are no fixed points in the given permutation. Also, the given sequence contains at least two numbers -1 and each positive number occurs in the sequence at most once. It's guaranteed that there is at least one suitable permutation.

Output

Output a single integer, the number of ways Iahub could recover his permutation, modulo 1000000007 (109 + 7).

Sample test(s)
input
```5
-1 -1 4 3 -1
```
output
```2
```
Note

For the first test example there are two permutations with no fixed points are [2, 5, 4, 3, 1] and [5, 1, 4, 3, 2]. Any other permutation would have at least one fixed point.

5
-1 -1 4 3 -1

5
-1 -1 2 5 -1

x 表示目前有多少个位置使得a[i] = -1 且数字 i 已经被填在某个位置上，称为无限制位置。

y 表示目前有多少个位置使得a[i] = -1 且数字 i 没有被填在某个位置上，称为有限制位置。

1）我们从 x 个无限制的的位置中找一个 j ，令a[i] = a[j]，a[j] = i。规约到d[i - 1]的方案数。

5
-1 -1 2 5 -1

2）我们从i - 1个有限制的位置中找一个位置 j ，令a[i] = j，a[j] = i。规约到d[i - 2]的方案数。

5
-1 -1 3 4 -1

3）我们从i - 1个有限制的位置中找一个位置 j ，令a[i] = j，但是a[j] != i。规约到d[i - 1]的方案数。也就相当于由原来的 d[j] != j 限制变为了d[j] != i，其它限制不变。

5
-1 -1 3 4 -1

AC代码：

```#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;
int mod=1e9+7;
__int64 d[2002];
int a[2002];

__int64 cal(int p)
{
__int64 ans=1;
int i;
for(i=2;i<=p;i++)
ans=(ans*i)%mod;
return ans;
}

int main()
{
int n,x,y;
while(~scanf("%d",&n))
{
int cnt=0,t=0;  //cnt记录可以填的个数,t记录填的数字被占用的个数
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
if(a[i]==-1)
cnt++;
}
for(int i=1;i<=n;i++)
{
if(a[i]>0)
{
if(a[a[i]]==-1)
t++;
}
}

x=t;  //x是无限制的个数
y=cnt-t;
d[0]=cal(x);   //d[0]=x!
for(int i=1;i<=y;i++)
{
d[i]=((x+i-1)*d[i-1])%mod;
if(i>1)
d[i]=(d[i]+(i-1)*d[i-2])%mod;
}
printf("%I64d\n",d[y]);
}
return 0;
}

/*
5
-1 -1 4 3 -1
5
-1 -1 4 -1 -1
5
-1 -1 2 5 -1
*/
```