Codeforces Round #114 (Div. 1) E. Wizards and Bets 高斯消元

E. Wizards and Bets

题目连接：

http://www.codeforces.com/contest/167/problem/E

Description

In some country live wizards. They like to make weird bets.

Two wizards draw an acyclic directed graph with n vertices and m edges (the graph's vertices are numbered from 1 to n). A source is a vertex with no incoming edges, and a sink is the vertex with no outgoing edges. Note that a vertex could be the sink and the source simultaneously. In the wizards' graph the number of the sinks and the sources is the same.

Wizards numbered the sources in the order of increasing numbers of the vertices from 1 to k. The sinks are numbered from 1 to k in the similar way.

To make a bet, they, as are real wizards, cast a spell, which selects a set of k paths from all sources to the sinks in such a way that no two paths intersect at the vertices. In this case, each sink has exactly one path going to it from exactly one source. Let's suppose that the i-th sink has a path going to it from the ai's source. Then let's call pair (i, j) an inversion if i < j and ai > aj. If the number of inversions among all possible pairs (i, j), such that (1 ≤ i < j ≤ k), is even, then the first wizard wins (the second one gives him one magic coin). Otherwise, the second wizard wins (he gets one magic coin from the first one).

Our wizards are captured with feverish excitement, so they kept choosing new paths again and again for so long that eventually they have chosen every possible set of paths for exactly once. The two sets of non-intersecting pathes are considered to be different, if and only if there is an edge, which lies at some path in one set and doesn't lie at any path of another set. To check their notes, they asked you to count the total winnings of the first player for all possible sets of paths modulo a prime number p.

Input

The first line contains three space-separated integers n, m, p (1 ≤ n ≤ 600, 0 ≤ m ≤ 105, 2 ≤ p ≤ 109 + 7). It is guaranteed that p is prime number.

Next m lines contain edges of the graph. Each line contains a pair of space-separated integers, ai bi — an edge from vertex ai to vertex bi. It is guaranteed that the graph is acyclic and that the graph contains the same number of sources and sinks. Please note that the graph can have multiple edges.

Output

Print the answer to the problem — the total winnings of the first player modulo a prime number p. Please note that the winnings may be negative, but the modulo residue must be non-negative (see the sample).

Sample Input

4 2 1000003

1 3

2 4

Sample Output

1

Hint

题意

给你一个有向无环图，然后保证这个图里面有k个点的入度为0，k个点的出度为0

然后现在你需要在这个图里面找k条不相交的路径，使得入度为0的点和出度为0的点一一对应

如果连接的逆序对数为偶数，那么得到一块钱，如果为奇数，就失去一块钱

现在问你所有的情况都找到之和，问你最后你有多少钱

钱需要mod一个数

题解：

假设我们不考虑相交这个条件

那么我们用m[i][j]表示从第i个入度为0的点到第j个出度为0的点的路径数量的话

显然最后的答案就是m这个矩阵的行列式值

这个东西高斯消元就好了

然后我们考虑相交这个东西，其实相交这个东西没什么卵用

假设有1-2,2-3,2-5,4-2这四条边的话

显然答案是0，但是我们考虑相交也没关系，因为正反就抵消了

1-2-3,4-2-5;1-2-5,4-2-3 这样子就抵消了

所以相交这个条件没什么用。

然后搞一搞就完了，这道题

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 650+5;
long long quickpow(long long  m,long long n,long long k)//返回m^n%k
{
    long long b = 1;
    while (n > 0)
    {
          if (n & 1)
             b = (b*m)%k;
          n = n >> 1 ;
          m = (m*m)%k;
    }
    return b;
}
int n,e,mod;
vector<int> E[maxn];
int in[maxn],out[maxn];
int f[maxn][maxn];
int m[maxn][maxn];
int q[maxn];
int S[maxn],T[maxn];
int k1=1,k2=1;
long long ans = 1;

void guess()
{
    for(int i=1;i<k1;i++)
    {
        if(m[i][i]==0)
        {
            for(int j=i+1;j<k1;j++)
            {
                if(m[j][i])
                {
                    for(int k=i;k<k1;k++)
                        m[i][k]=(m[i][k]+m[j][k])%mod;
                    break;
                }
            }
            if(m[i][i]==0)
            {
                puts("0");
                return;
            }
        }
        long long inv = quickpow(m[i][i],mod-2,mod);
        for(int j=i+1;j<k1;j++)
        {
            long long temp = m[j][i]*inv%mod;
            if(temp==0)continue;
            for(int k=i;k<k1;k++)
                m[j][k]=(m[j][k]-m[i][k]*temp)%mod;
        }
    }

    for(int i=1;i<k1;i++)
        ans = ans*m[i][i]%mod;
    if(ans<0)ans+=mod;
    cout<<ans<<endl;
}
int main()
{
    scanf("%d%d%d",&n,&e,&mod);
    for(int i=1;i<=e;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        in[y]++,out[x]++;
        E[x].push_back(y);
    }

    for(int i=1;i<=n;i++)
        if(in[i]==0)
            S[k1++]=i;
    for(int i=1;i<=n;i++)
        if(out[i]==0)
            T[k2++]=i;

    int l=1,r=1;
    for(int i=1;i<=n;i++)
        if(in[i]==0)
            q[r++]=i;
    while(l<r)
    {
        int now = q[l++];
        for(int i=0;i<E[now].size();i++)
        {
            in[E[now][i]]--;
            if(in[E[now][i]]==0)
                q[r++]=E[now][i];
        }
    }

    for(int i=1;i<k1;i++)
        f[S[i]][S[i]]=1;

    for(int i=1;i<=n;i++)
    {
        int u = q[i];
        for(int j=1;j<k1;j++)
        {
            int s = S[j];
            for(int k=0;k<E[u].size();k++)
                f[s][E[u][k]]=(f[s][E[u][k]]+f[s][u])%mod;
        }
    }

    for(int i=1;i<k1;i++)
        for(int j=1;j<k1;j++)
            m[i][j]=f[S[i]][T[j]];
    guess();

}