Bzoj 2169 Side--the thought of the heavy

Title: https://www.lydsy.com/JudgeOnline/problem.php?id=2169

If you have been to a good weight before, you can be seen as the edge of this connection will only repeat with the last one.

It can be thought that from the last state to this state, the process of transfer for each state of the last time is the remaining position of all the possibility of the edge of the traverse just once! That is, I am currently connected to the edge, and the number of repetitions of the last Edge is C (n,2)-(I-2) (n points in the selection of 2 is a total number of empty slots, the last time I-2 was placed, this time with the last can be repeated position there is so many).

With respect to the same scheme because of the different order of the number of repeated counting, as long as the one side after each count divided by I can be, for each scheme, the side of this time may be placed on the side of the I bar any one, resulting in repetition. It is not possible to do so, and finally divide by the factorial of the number of edges that are filled, because their transfer is not tenable when it is transferred from a repeating state.

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespacestd;Const intn=1005, mod=10007;intN,m,t,dp[n][n];BOOLDeg[n];intrdn () {intret=0;BOOLfx=1;CharCh=GetChar ();  while(ch>'9'|| ch<'0'){if(ch=='-') fx=0; ch=GetChar ();}  while(ch>='0'&&ch<='9') ret= (ret<<3) + (ret<<1) +ch-'0', ch=GetChar (); returnfx?ret:-ret;}intCalcintA) {return(A * (a)1) >>1)%MoD;}intpwintXintk) {  intret=1; while(k) {if(k&1) ret=ret*x%mod;x=x*x%mod;k>>=1;}returnret;}intMain () {n=rdn (); M=rdn (); t=RDN ();  for(intI=1, u,v;i<=m;i++) {u=rdn (); v=RDN (); Deg[u]=!deg[u]; deg[v]=!Deg[v]; }  intCnt=0;  for(intI=1; i<=n;i++) cnt+=Deg[i]; dp[0][cnt]=1;  for(intI=1; i<=t;i++)    {      intD=PW (i,mod-2);  for(intj=0; j<=n;j++) {Dp[i][j]=dp[i-1][j]*j%mod* (n-j)%MoD; DP[I][J]= (dp[i][j]+dp[i-1][j+2]*calc (j+2))%MoD; if(j>=2) Dp[i][j]= (dp[i][j]+dp[i-1][j-2]*calc (n-j+2))%MoD; if(i>1) dp[i][j]-=dp[i-2][j]* (Calc (n)-(i2))%MoD; if(dp[i][j]<0) dp[i][j]+=MoD; DP[I][J]=dp[i][j]*d%MoD; }} printf ("%d\n", dp[t][0]%MoD); return 0;}

Bzoj 2169 Side--the thought of the heavy