bzoj 2169 连边 —— DP+容斥

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
int const xn=1005,mod=10007;
int n,m,k,deg[xn];
ll f[xn][xn];
int rd()
{
  int ret=0,f=1; char ch=getchar();
  while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=0; ch=getchar();}
  while(ch>=‘0‘&&ch<=‘9‘)ret=(ret<<3)+(ret<<1)+ch-‘0‘,ch=getchar();
  return f?ret:-ret;
}
ll pw(ll a,int b)
{
  ll ret=1;
  for(;b;b>>=1,a=(a*a)%mod)
    if(b&1)ret=(ret*a)%mod;
  return ret;
}
ll calc(int x){return (ll)x*(x-1)/2;}
int main()
{
  n=rd(); m=rd(); k=rd(); int num=0;
  for(int i=1,x,y;i<=m;i++)x=rd(),y=rd(),deg[x]++,deg[y]++;
  for(int i=1;i<=n;i++)if(deg[i]&1)num++;
  f[0][num]=1;
  for(int i=1;i<=k;i++)
    for(int j=0;j<=n;j++)
      {
    f[i][j]=(f[i][j]+f[i-1][j]*((ll)j*(n-j)%mod)%mod)%mod;
    if(j>=2)f[i][j]=(f[i][j]+f[i-1][j-2]*calc(n-j+2)%mod)%mod;//+2!
    if(j<=n-2)f[i][j]=(f[i][j]+f[i-1][j+2]*calc(j+2)%mod)%mod;
    if(i>=2)f[i][j]=(f[i][j]-(f[i-2][j]*(calc(n)-i+2))%mod+mod)%mod;
    f[i][j]=(f[i][j]*pw(i,mod-2))%mod;//
      }
  printf("%lld\n",f[k][0]);
  return 0;
}