Bzoj 1017 Warcraft Map Dotr (tree dp)

Test instructions: There are two types of equipment, advanced equipment A and basic equipment B. Now there's M's money. Each b has a unit price and a maximum quantity that can be purchased. Each AI can be synthesized by other objects of CI, giving the number of other items required for CI species. Each outfit has a contribution value. The maximum contribution value. The synthetic route of known items is a strict tree model. There is one item that does not synthesize any other item, and the rest will only be used to synthesize another item.

Idea: F[i][j][k], representing the I-item, spent J, which offered K I goods.

1#include <algorithm>2#include <cstdio>3#include <cmath>4#include <cstring>5#include <iostream>6 intf[ -][2005][ the];7 intnum[20005],a[20005],b[20005],c[ the][ the],d[ the][ the];8 intn,m,vis[200005],dp[ -][2005],w[20005],pd[20005];9 voidDfsintx) {Ten     if(!Num[x]) { OneB[x]=std::min (b[x],m/a[x]); A          for(intI=0; i<=b[x];i++) -           for(intj=0; j<=i;j++) -f[x][i*a[x]][j]=w[x]* (I-j); the         return;  -     } -b[x]=0x7fffffff; -      for(intI=1; i<=num[x];i++){ +         intPur=C[x][i]; - dfs (PUR); +B[x]=std::min (b[x],b[pur]/d[x][i]); Aa[x]+=d[x][i]*A[pur]; at     } -B[x]=std::min (b[x],m/a[x]); -      for(intt=0; t<=b[x];t++){ -memset (dp,-0x3f3f3f3f,sizeofDP); -dp[0][0]=0;  -          for(intI=1; i<=num[x];i++) in           for(intj=0; j<=m;j++) -            for(intk=0; k<=j;k++) toDp[i][j]=std::max (dp[i][j],dp[i-1][j-k]+f[c[x][i]][k][t*D[x][i]]); +          for(intj=0; j<=m;j++) -           for(intk=0; k<=t;k++) theF[x][j][k]=std::max (f[x][j][k],dp[num[x]][j]+ (t-k) *w[x]); *     } $ }Panax Notoginseng intMain () { -     Chars[ -]; thescanf"%d%d",&n,&m); +      for(intI=1; i<=n;i++){ Ascanf"%d",&w[i]); thescanf"%s", s+1); +         if(s[1]=='A') pd[i]=1;Elsepd[i]=0; -num[i]=0; $         if(pd[i]==0) { $scanf"%d%d",&a[i],&b[i]); -num[i]=0; -           Continue; the         } -scanf"%d",&num[i]);Wuyi          for(intj=1; j<=num[i];j++){ thescanf"%d%d",&c[i][j],&d[i][j]); -vis[c[i][j]]=1;  Wu         } -     } About     inti; $      for(i=1; i<=n;i++)if(!vis[i]) Break; -     introot=i; -memset (f,-0x3f3f3f3f,sizeoff); - Dfs (root); A     intans=0; +      for(intI=0; i<=m;i++) the       for(intj=0; j<=b[root];j++) -ans=Std::max (Ans,f[root][i][j]); $printf"%d\n", ans);  the}

Bzoj 1017 Warcraft Map Dotr (tree dp)