bzoj4867: [Ynoi2017]舌尖上的由乃
作者：互联网
没得去冬令营的菜鸡只能在机房刷题耍子咯
这道题需要平衡块的大小，好像thuwc d1t1就考了
先分块，设块的大小为k，每一块块内排序，复杂度O(nlogk)
修改完整块打标记，不完整暴力，复杂度O(n/k+k)
询问二分答案，整块再二分位置，不完整的暴力拿出来，也二分，复杂度O(logn*n/k*logk+k)
明显最后一个比较大，可以让logn*n/k*logk=k，这样会比较平衡
logk约等于logn，于是块的大小设为sqrt(n)*logn
实测k=sqrt(n)，询问强的要跑25s+,弱的1s
k=sqrt(n)*logn就都是5s了
分块细节非常多啊。。。。。。码量也很大
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn=110000;
inline int read()
{
int x=0,f=1;char 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(int d)
{
if(d<0){putchar('-');d=-d;}
if(d>=10)write(d/10);
putchar(d%10+'0');
}
struct node
{
int x,y,d,next;
}a[maxn*2];int len,last[maxn];
inline void ins(int x,int y,int d)
{
len++;
a[len].x=x;a[len].y=y;a[len].d=d;
a[len].next=last[x];last[x]=len;
}
struct point{int d,id;}p[maxn];int plen;
inline bool cmp(point p1,point p2){return p1.d==p2.d?p1.id<p2.id:p1.d<p2.d;}
int L[maxn],R[maxn];
void dfs(int x)
{
int now=++plen; p[now].id=plen;
L[x]=plen;
for(int k=last[x];k;k=a[k].next)
{
int y=a[k].y;
p[plen+1].d=p[now].d+a[k].d;
dfs(y);
}
R[x]=plen;
}
//-------------------------------------------init------------------------------------------------------
int n,block,st[maxn],ad[maxn]; int op,x,k; int Lli,Rli;
inline int be(int tt){return (tt-1)*block+1;}
inline bool inin(int i){return (L[x]<=p[i].id&&p[i].id<=R[x]);}
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
point t1[maxn],t2[maxn];int t1len,t2len;
inline void mergesort(int u)
{
int i=1,j=1,o=be(u);
while(i<=t1len&&j<=t2len)
if(cmp(t1[i],t2[j]))p[o++]=t1[i++];
else p[o++]=t2[j++];
while(i<=t1len)p[o++]=t1[i++];
while(j<=t2len)p[o++]=t2[j++];
}
inline void bl_change(int u)
{
Lli=be(u),Rli=min(be(u+1)-1,n);
t1len=0;t2len=0;
for(int i=Lli;i<=Rli;i++)
if(inin(i))p[i].d+=k,t1[++t1len]=p[i];
else t2[++t2len]=p[i];
mergesort(u);
}
//~~~~~~~~~~~~change~~~~~~~~~~~~~~~~~~~~
inline int bl_findkth(int u)
{
Lli=be(u),Rli=min(be(u+1)-1,n);
for(int i=Lli;i<=Rli;i++)
if(inin(i))
{
k--;
if(k==0)return p[i].d+ad[u];
}
}
inline bool check(int D,int u,int v)
{
int ret=0;
int l,r,num;
l=1,r=t1len,num=0; while(l<=r){ int mid=(l+r)/2; if(D>=t1[mid].d+ad[u])l=mid+1,num=mid; else r=mid-1; } ret+=num;
l=1,r=t2len,num=0; while(l<=r){ int mid=(l+r)/2; if(D>=t2[mid].d+ad[v])l=mid+1,num=mid; else r=mid-1; } ret+=num;
if(ret>=k)return true;
for(int i=u+1;i<v;i++)
{
int l=be(i),r=min(be(i+1)-1,n),num=l-1;
while(l<=r)
{
int mid=(l+r)/2;
if(D>=p[mid].d+ad[i])l=mid+1,num=mid;
else r=mid-1;
}
ret+=num-be(i)+1;
if(ret>=k)return true;
}
return false;
}
//~~~~~~~~~~~~~getans~~~~~~~~~~~~~~~~~~~~
//-----------------------------------------------solve--------------------------------------------------------
int main()
{
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
int Q,F,dd;
n=read(),Q=read(),read();
for(int i=2;i<=n;i++)
F=read(),dd=read(),ins(F,i,dd);
dfs(1);
block=int(sqrt(double(n+1))*log(n)/log(2));
int h=1;
for(int i=1;i<=n;i++)
{
st[i]=(i-1)/block+1;
if(i!=1&&st[i]!=st[i-1])
sort(p+h,p+i,cmp),h=i;
}
sort(p+h,p+n+1,cmp);
while(Q--)
{
op=read(),x=read(),k=read();
if(op==2)
{
int u=st[L[x]],v=st[R[x]];
if(u==v)bl_change(u);
else
{
bl_change(u),bl_change(v);
for(int i=u+1;i<v;i++)ad[i]+=k;
}
}
else
{
if(R[x]-L[x]+1<k){puts("-1");continue;}
int u=st[L[x]],v=st[R[x]];
if(u==v)write(bl_findkth(u)),puts("");
else
{
t1len=0;
Lli=be(u),Rli=min(be(u+1)-1,n);
for(int i=Lli;i<=Rli;i++)
if(inin(i))t1[++t1len]=p[i];
t2len=0;
Lli=be(v),Rli=min(be(v+1)-1,n);
for(int i=Lli;i<=Rli;i++)
if(inin(i))t2[++t2len]=p[i];
int l=1,r=1e7,ans;
while(l<=r)
{
int mid=(l+r)/2;
if(check(mid,u,v))r=mid-1,ans=mid;
else l=mid+1;
}
write(ans);puts("");
}
}
}
return 0;
}
标签：ch,bzoj4867,int,mid,len,Ynoi2017,舌尖,num,include 来源： https://www.cnblogs.com/AKCqhzdy/p/10301822.html