[线段树 均摊复杂度] BZOJ 2130 魔塔

不会做QAQ 还是转身向Claris的题解吧

#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;}