[線段樹 均攤複雜度] BZOJ 2130 魔塔
阿新 • • 發佈:2019-01-11
#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline void read(int &x){
char c=nc(),b=1;
for (;!(c>='0' && c<='9' );c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}
const int N=100005;
int F[N<<2];
int vm[N<<2],lm[N<<2],rm[N<<2];
int n,K[N],a[N],b[N],c[N],sa[N],sb[N],sc[N],pb[N],pc[N];
int f[N];
inline void upd(int x){
lm[x]=lm[x<<1]; rm[x]=rm[x<<1 |1]; vm[x]=max(vm[x<<1],vm[x<<1|1]);
}
inline void Build(int x,int l,int r){
F[x]=-1;
if (l==r){
lm[x]=rm[x]=f[l],vm[x]=f[l]+sb[l];
return;
}
int mid=(l+r)>>1;
Build(x<<1,l,mid); Build(x<<1|1,mid+1,r);
upd(x);
}
inline void mark(int x,int r,int t){
F[x]=lm[x]=rm[x]=t; vm[x]=t+sb[r];
}
inline void Modify(int x,int l,int r,int ql,int qr,int t){
if (ql<=l && r<=qr){
if (lm[x]<=t) return;
if (rm[x]>=t) { mark(x,r,t); return; }
}
int mid=(l+r)>>1;
if (F[x]!=-1) mark(x<<1,mid,F[x]),mark(x<<1|1,r,F[x]),F[x]=-1;
if (ql<=mid) Modify(x<<1,l,mid,ql,qr,t);
if (qr>mid) Modify(x<<1|1,mid+1,r,ql,qr,t);
upd(x);
}
int main(){
int T;
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
read(T);
while (T--){
read(n);
for (int i=1;i<=n;i++) read(K[i]);
for (int i=1;i<=n;i++) read(a[i]);
for (int i=1;i<=n;i++) read(b[i]),pb[b[i]]=i;
for (int i=1;i<=n;i++) read(c[i]),pc[c[i]]=i;
for (int i=1;i<=n;i++) read(sa[i]),sa[i]+=sa[i-1];
for (int i=1;i<=n;i++) read(sb[i]),sb[i]+=sb[i-1];
for (int i=1;i<=n;i++) read(sc[i]),sc[i]+=sc[i-1];
f[0]=sc[n];
for (int i=1;i<=n;i++)
if (K[b[i]]==1)
f[i]=min(f[i-1],sc[pc[b[i]]-1]);
else
f[i]=f[i-1];
Build(1,0,n);
int ans=vm[1];
for (int i=1;i<=n;i++){
if (K[a[i]]==1){
Modify(1,0,n,pb[a[i]],n,-1<<30);
Modify(1,0,n,0,n,sc[pc[a[i]]-1]);
}else{
Modify(1,0,n,pb[a[i]],n,sc[pc[a[i]]-1]);
}
ans=max(ans,sa[i]+vm[1]);
}
printf("%d\n",ans);
}
return 0;
}