BZOJ2169 连边（动态规划）

　　令f[i][j]表示连i条边时奇点个数为j的方案数，转移时讨论两奇点相连、一奇一偶相连、两偶点相连即可。注意这样会造成重边，那么算出恰好有一条重边的方案数并减掉。由于是有序地考虑每条边，每次还要除以i。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 1010
#define P 10007
int n,m,k,degree[N],f[N][N],inv[N],ans,cnt;
int C(int n,int m){return (n*(n-)>>)%P;}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj2169.in","r",stdin);
    freopen("bzoj2169.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read(),k=read();
    for (int i=;i<=m;i++)
    {
        int x=read(),y=read();
        degree[x]^=,degree[y]^=;
    }
    for (int i=;i<=n;i++) if (degree[i]) cnt++;
    inv[]=;inv[]=;for (int i=;i<=k;i++) inv[i]=P-(P/i)*inv[P%i]%P;
    f[][cnt]=;
    for (int i=;i<=k;i++)
        for (int j=;j<=n;j++)
        f[i][j]=(f[i-][j+]*C(j+,)%P+f[i-][j]*j%P*(n-j)%P+(j>=?f[i-][j-]*C(n-j+,)%P:)-(i>=?f[i-][j]*(C(n,)-i+)%P:)+P)%P*inv[i]%P;
    cout<<f[k][];
    return ;
}