Cogs 444. [HAOI2010] Software Installation

★★☆ input file: install.in output file: install.out Simple comparison
Time limit: 1 s memory limit: MB

"Problem description"
Now we have n software on hand, for a software I, it will occupy WI disk space, its value is VI. We want to choose from some software installed on a disk capacity of M computer, so that the value of these software as large as possible (that is, VI and the largest).
But now there is a problem: there is a dependency between software, that is, software I can work correctly only if software J (including Software J's direct or indirect dependency) is installed (software I relies on software j). Fortunately, one software relies on another software. If a software does not work properly, then it can play a role of 0.
We now know the dependencies between software: Software I relies on software di. Now, please design a solution to install the software as large as possible. A software can only be installed once, if a software is not dependent on the di=0, then as long as the software installed, it will work properly.

"Input Format"

Line 1th: N,m (0<=n<=100. 0<=M<=500)
Line 2nd: w1,w2,...,wi,...,wn (0<=wi<=m)
Line 3rd: V1,V2,...,VI,...,VN (0<=vi<=1000)
Line 4th: D1,d2,...,di,...,dn (0<=di<=n,di≠i)

"Output Format"

An integer that represents the maximum value.

"Input Sample"
3 10
5 5 6
2 3 4
0 1 1

"Output Example"
5

Exercises

　　Depending on the dependency, you can draw a picture, there are three possible scenarios: 1. Dependencies form a tree 2. Dependencies constitute a ring 3. Dependencies form a ring under which a tree is hanging. Because there are 2, 3 of these conditions, so we have to tarjan pre-treatment, contraction ring for the point, re-build the map.

For the built diagram, run one side of the tree backpack, thinking similar to 01 backpack, F[x][tot] for the root of x, the capacity of TOT is the biggest benefit. Consider each subtree of X as an item, and then enumerate the capacity assigned to each subtree, and tot shifts from large to small.

Also a little, f[x][tot] to ensure that X to count in, the last processing can be guaranteed.

1#include <iostream>2#include <cstdio>3#include <cstdlib>4#include <cstring>5#include <cmath>6#include <algorithm>7#include <queue>8#include <vector>9 using namespacestd;Ten Const intmaxn= $, maxm= -; One intN,M,W[MAXN],V[MAXN],FA[MAXN]; A intVAL[MAXN],COST[MAXN],F[MAXN][MAXM]; -  -vector<int>TO[MAXN],SON[MAXN]; the intstac[maxn],top=0, Dfn[maxn],low[maxn],inkin[maxn],tot,index; - BOOLINSTAC[MAXN]; -vector<int>KIN[MAXN]; -InlinevoidTarjan (intx) { +dfn[x]=low[x]=index++; -stac[++top]=x; +instac[x]=true; A      for(intI=0; I<to[x].size (); i++){ at         inty=To[x][i]; -         if(dfn[y]==-1){ - Tarjan (y); -low[x]=min (low[x],low[y]); -         } -         Else if(instac[y]!=0){ inlow[x]=min (low[x],dfn[y]); -         } to     } +     if(dfn[x]==Low[x]) { -tot++; the         inty; *          Do{ $y=stac[top--];Panax Notoginsenginstac[y]=false; - kin[tot].push_back (y); theinkin[y]=tot; +} while(y!=x); A     } the } +InlinevoidCalcintx) { -     if(son[x].size () = =0){ $          for(inti=cost[x];i<=m;i++) f[x][i]=Val[x]; $         return ; -     } -      for(intI=0; I<son[x].size (); i++) Calc (son[x][i]); the      -      for(intI=0; I<son[x].size (); i++){Wuyi         inty=Son[x][i]; the          for(inttot=m;tot>=0; tot--){ -              for(intj=0; j<=tot;j++){ WuF[x][tot]=max (f[x][tot],f[x][tot-j]+f[y][j]); -             } About         } $     } -      for(inti=m;i>=0; i--){ -         if(I>=cost[x]) f[x][i]=f[x][i-cost[x]]+Val[x]; -         Elsef[x][i]=0; A     } + } the intMain () { -scanf"%d%d",&n,&M); $      for(intI=1; i<=n;i++) scanf ("%d",&w[i]); the      for(intI=1; i<=n;i++) scanf ("%d",&v[i]); the      for(intI=1; i<=n;i++){ thescanf"%d",&fa[i]); the To[fa[i]].push_back (i); -     } inmemset (dfn,-1,sizeof(DFN)); the      for(intI=1; i<=n;i++){ the         if(dfn[i]==-1) Tarjan (i); About     } the      for(intI=1; i<=tot;i++){ the         inty=kin[i][0]; the         if(Kin[i].size () >=2){//form a ring, take any one of them, and indent the ring as a point +Val[y]=v[y]; cost[y]=W[y]; -son[0].push_back (y); the              for(intj=1; J<kin[i].size (); j + +){Bayival[y]+=V[kin[i][j]]; thecost[y]+=W[kin[i][j]]; the             } -         } -         Else{//is a certain point in a tree, copied directly the             if(fa[y]==0)  theson[0].push_back (y); the             Else{ the                 intxx=Inkin[fa[y]]; -                 intyy=kin[xx][0]; the son[yy].push_back (y); the             } theVal[y]=v[y]; cost[y]=W[y];94         } the     } theval[0]=0; cost[0]=0; Calc0); theprintf"%d", f[0][m]);98     return 0; About}

Cogs 444. [HAOI2010] Software Installation