[历史最值线段树] UOJ#164. 【清华集训2015】V

这种标记很强啊…
UR#11 C题解

#include <cstdio>#include <algorithm>#include <iostream>using namespace std;typedef long long ll;const int N=500010;const ll inf=1LL<<60;int n,m,a[N];struct tag{  ll a,b;  tag(ll x=0,ll y=-inf):a(x),b(y){}  friend tag operator +(tag b,tag a){    return tag(max(a.a+b.a,-inf),max(a.b,a.a+b.b));  }  bool empty(){    return !a && b==-inf;  }}F[N<<2],H[N<<2],t[N<<2];inline tag max(tag a,tag b){  return tag(max(a.a,b.a),max(a.b,b.b));}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(); x=0;  for(;c>'9' || c<'0';c=nc());for(;c>='0'&&c<='9';x=x*10+c-'0',c=nc());}inline void add(int g,tag f,tag h){  H[g]=max(H[g],F[g]+h); F[g]=F[g]+f;}inline void Push(int g){  if(F[g].empty()) return ;  add(g<<1,F[g],H[g]); add(g<<1|1,F[g],H[g]);  F[g]=H[g]=tag();}inline void Add(int g,int l,int r,int L,int R,tag x){  if(l==L && r==R) return add(g,x,x);  int mid=L+R>>1; Push(g);  if(r<=mid) Add(g<<1,l,r,L,mid,x);  else if(l>mid) Add(g<<1|1,l,r,mid+1,R,x);  else Add(g<<1,l,mid,L,mid,x),Add(g<<1|1,mid+1,r,mid+1,R,x);}inline int Pushtag(int g,int x,int L,int R){  if(L==R) return g;  int mid=L+R>>1; Push(g);  if(x<=mid) Pushtag(g<<1,x,L,mid); else Pushtag(g<<1|1,x,mid+1,R);}void PutAns(ll x){  if(x>=10) PutAns(x/10); putchar(x%10+'0');}int main(){  freopen("1.in","r",stdin);  freopen("1.out","w",stdout);  read(n); read(m);  for(int i=1;i<=n;i++) read(a[i]);  while(m--){    int opt,l,r,x; read(opt);    if(opt==1){      read(l); read(r); read(x);      Add(1,l,r,1,n,tag(x,-inf));    }    else if(opt==2){      read(l); read(r); read(x);      Add(1,l,r,1,n,tag(-x,0));    }    else if(opt==3){      read(l); read(r); read(x);      Add(1,l,r,1,n,tag(-inf,x));    }    else if(opt==4){      read(x);       int c=Pushtag(1,x,1,n);      PutAns(max(a[x]+F[c].a,F[c].b)); putchar('\n');    }    else{      read(x);       int c=Pushtag(1,x,1,n);      PutAns(max(a[x]+H[c].a,H[c].b)); putchar('\n');    }  }  return 0;}