bzoj2427 [HAOI2010] Software Installation

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

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

An integer that represents the maximum value.

Sample INPUT3 10
5 5 6
2 3 4
0 1 1 Sample Output5

Tarjan Shrink Point Tree backpack DP

Huang Huge said this kind of problem a lot but I am the first time to do ah ...

#include <cstdio> #include <iostream> #include <cstring> #include <cstdlib> #include < algorithm> #include <cmath> #include <queue> #include <deque> #include <set> #include <map > #include <ctime> #define LL long long#define INF 0x7ffffff#define pa pair<int,int> #define Pi 3.1415926535897932384626433832795028841971#define N 110#define M 510using namespace Std;inline ll read () {ll x=0,f=1;ch    Ar Ch=getchar (); while (ch< ' 0 ' | |    Ch> ' 9 ') {if (ch== '-') F=-1;ch=getchar ();}    while (ch>= ' 0 ' &&ch<= ' 9 ') {x=x*10+ch-' 0 '; Ch=getchar ();} return x*f;}    inline void Write (LL a) {if (a<0) {printf ("-"); a=-a;}    if (a>=10) write (A/10); Putchar (a%10+ ' 0 ');} inline void Writeln (LL a) {write (a);p rintf ("\ n");} struct Edge{int to,next;} E1[2*n],e2[2*n];int head1[n],head2[n];int c[n],w[n],fa[n],belong[n];int c[n],w[n],f[n][m];int dfn[N],low[N];int Zhan[n],top;bool inset[n],mrk[n];int size[n];int n,m,cnt1,cnt2,cnt3,tt,ans;inline void ins1 (int u, int v) {e1[++cnt1].to=v;    E1[cnt1].next=head1[u]; Head1[u]=cnt1;}    inline void ins2 (int u,int v) {e2[++cnt2].to=v;    E2[cnt2].next=head2[u]; Head2[u]=cnt2;}    inline void dfs (int x) {low[x]=dfn[x]=++tt;    Zhan[++top]=x;inset[x]=1;            for (int i=head1[x];i;i=e1[i].next) {if (!dfn[e1[i].to]) {DFS (e1[i].to);        Low[x]=min (low[x],low[e1[i].to]);    }else if (inset[e1[i].to]) low[x]=min (low[x],dfn[e1[i].to]);        } if (Dfn[x]==low[x]) {cnt3++;        int p=-1;            while (p!=x) {p=zhan[top--];            inset[p]=0;            Belong[p]=cnt3;            C[CNT3]+=C[P];        W[CNT3]+=W[P]; }}}inline void Tarjan () {for (int i=1;i<=n;i++) if (!dfn[i]) DFS (i);}    inline void treedp (int x) {if (mrk[x]) return;mrk[x]=1;        for (int i=head2[x];i;i=e2[i].next) if (!mrk[e2[i].to]) {TREEDP (e2[i].to); for (int j=m-c[x];j>=0;j--) for (int k=j; k>=0;k--) F[x][j]=max (F[x][j],f[x][j-k]+f[e2[i].to][k]);        } for (int i=m;i>=0;i--) {if (i>=c[x]) f[x][i]=f[x][i-c[x]]+w[x];    else f[x][i]=0;    }}int Main () {n=read (); M=read ();    for (int i=1;i<=n;i++) c[i]=read ();    for (int i=1;i<=n;i++) w[i]=read ();        for (int i=1;i<=n;i++) {fa[i]=read ();    if (Fa[i]) ins1 (fa[i],i);    } Tarjan (); for (int i=1;i<=n;i++) {for (int j=head1[i];j;j=e1[j].next) if (Belong[e1[j].to]!=belong[i]) ins2 (Belong[i],    Belong[e1[j].to]);    } for (int i=1;i<=cnt3;i++) ins2 (0,i);    TREEDP (0);    for (int i=0;i<=m;i++) Ans=max (Ans,f[0][i]); printf ("%d\n", ans);}

　　

bzoj2427 [HAOI2010] Software Installation