Codeforces Round #277.5(Div. 2) F. Special Matrices【思维+Dp】好题~好题~

F. Special Matrices

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

An n × n square matrix is
special, if:

it is binary, that is, each cell contains either a 0, or a 1;
the number of ones in each row and column equals 2.
You are given n and the first
m rows of the matrix. Print the number of special
n × n matrices, such that the first
m rows coincide with the given ones.

As the required value can be rather large, print the remainder after dividing the value by the given number
mod.

Input
The first line of the input contains three integers n,
m, mod (2 ≤ n ≤ 500,
0 ≤ m ≤ n,
2 ≤ mod ≤ 109). Then
m lines follow, each of them contains n characters — the first rows of the required special matrices. Each of these lines contains exactly two characters '1', the
rest characters are '0'. Each column of the given
m × n table contains at most two numbers one.

Output
Print the remainder after dividing the required value by number
mod.

Examples

Input

3 1 1000
011

Output

2

Input

4 4 100500
0110
1010
0101
1001

Output

1

Note
For the first test the required matrices are:

011101110

011110
101

In the second test the required matrix is already fully given, so the answer is 1.

题目大意：

让你构造出来一个N*N的矩阵，使得每行每列保证只有两个1。

现在已经给出M行的矩阵部分，问最终可以得到多少个可行方案。

思路（思路源自：http://blog.csdn.net/u014733623/article/details/41250593）：

1、一开始想的是设定dp【i】【x】【y】【2（k）】【2（l）】表示到第i列 ，这一列的放置1的两个位子在x，y上，对应这两行上边出现1的个数为k，l个的方案数。

状态转移方程很好写，但是时间复杂度不允许，会达到O（n^5*4）；

再之后就不会设定了。

其实反向想一下，考虑剩余部分就好了。

2、设定dp【s1】【s2】表示到第i行，剩余s1个列可以放置两个1，剩余s2个列可以放置1个1的方案数。

那么就有：

dp【s1-2】【s2】=dp【s1】【s2】*C(s1,2)；

dp【s1-1】【s2】=dp【s1】【s2】*C(s1,1)*C（s2,1）；

dp【s1】【s2-2】=dp【s1】【s2】*C(s2,2)；

我们可以枚举行和s1，s2=（n-i+1）*2-s1*2【剩余需要放置1的个数-s1*2==剩余需要放置1的个数==s2】；

时间复杂度O（n^2）；

解法真的很巧妙。点个赞。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<vector>
using namespace std;
#define ll __int64
char a[505][505];
ll dp[606][606];
ll num[606];
int main()
{
ll n,m,mod;
while(~scanf("%I64d%I64d%I64d",&n,&m,&mod))
{
memset(dp,0,sizeof(dp));
memset(num,0,sizeof(num));
for(ll i=1;i<=m;i++)scanf("%s",a[i]+1);
ll s1=0;
ll s2=0;
for(ll i=1;i<=m;i++)
{
for(ll j=1;j<=n;j++)
{
if(a[i][j]=='1')num[j]++;
}
}
for(ll j=1;j<=n;j++)
{
if(num[j]==0)s1++;
if(num[j]==1)s2++;
}
dp[s1][s2]=1;
for(ll i=m+1;i<=n;i++)
{
for(ll s1=0;s1<=n;s1++)
{
ll s2=(n-i+1)*2-s1*2;
if(s2<0||s2>n)continue;
if(s1-2>=0)dp[s1-2][s2+2]+=(dp[s1][s2]*(s1)*(s1-1)/2)%mod;
if(s1-1>=0)dp[s1-1][s2]+=(dp[s1][s2]*(s1)*(s2))%mod;
if(s2-2>=0)dp[s1][s2-2]+=(dp[s1][s2]*(s2)*(s2-1)/2)%mod;
}
}
printf("%I64d\n",dp[0][0]);
}
}