hdu4779 Combination Count +DP

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Test instructions: Given the n*m grid, then a P-heavy defense tower, capable of withstanding 1 attacks, q a light tower can not accept any attack, then each defense will attack his row and the column, and finally to put at least one defense tower on this grid program number,

We enumerate how many heavy defense towers are selected and how much of this heavy defense tower is two in one row, and two in a row O (p^3) efficiency

#include <iostream>#include<algorithm>#include<cstdio>#include<string.h>#include<vector>using namespaceStd;typedefLong LongLL;ConstLL mod=1000000007;Const intmaxn=202; LL C[MAXN][MAXN];//Number of combinationsLL LIGHT[MAXN][MAXN][MAXN];//light[i][j][k]k stack Lightweight defense tower put in the I row J column of the Matrix scheme numberLL HEAVY[MAXN];//Haeavy[i] Represents the number of programs with a weight of two of the same rowLL NT[MAXN];//n! N factorialvoidinit () {nt[0]=1;  for(LL i=1; i<= $; i++) Nt[i]= (nt[i-1]*i)%MOD; memset (C,0,sizeof(C)); c[0][0]=1;  for(intI=1; i<= $; i++) {c[i][0]=1;  for(intj=1; j<=i; J + +) C[i][j]= (c[i-1][j]+c[i-1][j-1])%MOD; }     for(intI=1; i<= $; i++)         for(intj=1; j<= $; J + +)        {             intkm=min (i,j); light[I [j] [0] =1;  for(intk=1; k<=km; k++) {light[i] [j] [K]= ((((c[i] [k]*c[J] [K])%mod) *nt[k])%MOD; }        }     for(intI=1; i<= $; i++)         for(intj=1; j<= $; J + +)        {             intkm=min (i,j);  for(intk=1; k<=km; k++) Light[i][j][k]= (light[i][j][k-1]+LIGHT[I][J][K])%MOD; } heavy[0]=1;  for(intI=1; i<= -; i++) {LL ans=1;  for(intj=i*2; J>2; j-=2) {ans= (c[j][2]*ans)%MOD; } Heavy[i]=ans; }}intMain () {init (); intn,m,p,q; intCAs; scanf ("%d",&CAs);  for(intCc=1; cc<=cas; Cc++) {scanf ("%d%d%d%d",&n,&m,&p,&Q); LL ans=0;  for(intk=1; k<=p; k++)             for(intI=0; i<=k; i+=2)               for(intj=0; j+i<=k; j+=2)              {                 intln=n-i/2-J; intlm=m-j/2-i; if(min (LN,LM) <k-(i+j))Continue; LN=n,lm=l; LL D=1; D= (((c[ln][i/2]*c[lm][i]%mod) *heavy[i/2])%MOD; LN-=i/2; lm-=i; D= (d* (((c[ln][j) * c[lm][j/2])%mod) * heavy[j/2]) (%MOD))%MOD; LN-=J; LM-=j/2; intlest=k-(i+j); D= ((((D* ((c[ln][lest]*c[lm][lest)%mod)%mod) *nt[lest])%MOD; LN-=lest; lm-=lest; if(ln>0&&LM>0)                 {                     intGe=min (min (ln,lm), Q); D= (d* (light[ln][lm][ge))%MOD; } ans= (ans+d)%MOD; }        if(q>0)        {            intGe=min (min (n,m), Q); Ans= (Ans+light[n][m][ge])%MOD; Ans= (ans-1+MOD)%MOD; } printf ("%i64d\n", ans%MOD); }    return 0;}

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hdu4779 combined Count +DP