文章目錄

• 題意
• 題解

題意

$一個長為n的序列,一開始全為2.每次操作將一個區間賦值為1或2,並詢問整體有多少個2.$

題解

可以用權值線段樹瞎搞. 當然用某暴力資料結構珂朵莉樹也可以過啦. 每次動態維護答案,用$set$合併區間,暴力拆開. 然後就過了. 珂朵莉樹的話全部都是基本操作,具體可以看洛谷上CF896C的題解.

#include<bits/stdc++.h>
 
 //Ithea Myse Valgulious
namespace chtholly{
typedef long long ll;
#define re0 register int
#define rel register ll
#define rec register char
#define gc getchar
//#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<23,stdin),p1==p2)?-1:*p1++)
#define pc putchar
#define p32 pc(' ')
#define pl puts("")
 

/*By Citrus*/
char buf[1<<23],*p1=buf,*p2=buf;
inline int read(){
  int x=0,f=1;char c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return f?x:-x;
  }
template <typename mitsuha>
inline bool read(mitsuha &x){
  x=0;int f=1
 
;char c=gc();
  for (;!isdigit(c)&&~c;c=gc()) f^=c=='-';
  if (!~c) return 0;
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return x=f?x:-x,1;
  }
template <typename mitsuha>
inline int write(mitsuha x){
  if (!x) return 0&pc(48);
  if (x<0) pc('-'),x=-x;
  int bit[20],i,p=0;
  for (;x;x/=10) bit[++p]=x%10;
  for (i=p;i;--i) pc(bit[i]+48);
  return 0;
  }
inline char fuhao(){
  char c=gc();
  for (;isspace(c);c=gc());
  return c;
  }
}using namespace chtholly;
using namespace std;

struct chtholly_tree{
  #define it set<node>::iterator
  int sum;
  struct node{
    int l,r,val;
    node(int nl,int nr=-1,int v=0):l(nl),r(nr),val(v){}
    bool operator <(const node &b) const{
      return l<b.l;
    } 
  };
  set<node> s;

  it split(int pos) {
    it p=s.lower_bound(node(pos));
    if (p!=s.end()&&p->l==pos) return p; --p;
    int nl=p->l,nr=p->r,v=p->val;
    s.erase(p);
    s.insert(node(nl,pos-1,v));
    return s.insert(node(pos,nr,v)).first;
  }
  
  void assign(int l,int r,int v) {
    it p,itr=split(r+1),itl=split(l);
    for (p=itl;p!=itr;++p) sum-=(p->r-p->l+1)*p->val;
    s.erase(itl,itr);
    s.insert(node(l,r,v)),sum+=(r-l+1)*v;
  }

  void init(int n) {
    sum=n;
    s.insert(node(1,n,1));
  }
}my_;

int main(){
  int n=read(),q=read();
  my_.init(n);
  for (;q--;) {
    int l=read(),r=read(),k=read()-1;
    my_.assign(l,r,k),write(my_.sum),pl;
  }
}

謝謝大家.