题目链接:https://www.luogu.com.cn/problem/P2572
题目大意:
思路:珂朵莉树搞一搞,好像数据加强了,只能30分,会T。
#include <bits/stdc++.h> #define ll long long #define sit set<node>::iterator using namespace std; struct node{ int l, r; mutable ll val; int operator<(const node &a) const{ return l<a.l; } node(int L, int R, ll Val): l(L), r(R), val(Val){} node(int L) : l(L){} }; struct odTree{ set<node> s; sit Split(int pos){//一个节点分裂[l, r]->[l, pos-1]+[pos, r] sit it=s.lower_bound(node(pos)); if(it!=s.end()&&it->l==pos) return it; --it; int l=it->l, r=it->r; ll val=it->val; s.erase(it); s.insert(node(l, pos-1, val)); return s.insert(node(pos, r, val)).first; } void Assign(int l, int r, ll val){//区间赋值 sit it2=Split(r+1), it1=Split(l); s.erase(it1, it2); s.insert(node{l, r, val}); } void Add(int l, int r){//区间[l, r] a[i]取反 sit it2=Split(r+1), it1=Split(l); for(sit it=it1; it!=it2; ++it) it->val^=1; } ll Query(int l, int r){//求[l, r]的1的个数 sit it2=Split(r+1), it1=Split(l); ll res=0; for(sit it=it1; it!=it2; ++it){ if(it->val==1){ res+=((it->r)-(it->l)+1); } } return res; } ll Query2(int l, int r){//求[l, r]的0连续1的个数 sit it2=Split(r+1), it1=Split(l); ll res=0, tmp=0; for(sit it=it1; it!=it2; ){ if(it->val==1){ tmp=0; while(it!=it2&&it->val==1){ tmp+=((it->r)-(it->l)+1); ++it; } res=max(res, tmp); } else{ ++it; } } return res; } }Tree; 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<<1)+(x<<3)+(ch^48); ch=getchar(); } return x*f; } int main(){ int n, m, val; n=read(), m=read(); for(int i=1; i<=n; i++){ val=read(); Tree.s.insert(node(i, i, (ll)val)); } Tree.s.insert(node(n+1, n+1, -1)); for(int i=1; i<=m; i++){ int op=read(), x=read(), y=read(); x++, y++; if (op == 0) Tree.Assign(x, y, 0); else if (op == 1) Tree.Assign(x, y, 1); else if (op == 2) Tree.Add(x, y); else if (op == 3) printf("%lld\n", Tree.Query(x, y)); else printf("%lld\n", Tree.Query2(x, y)); } return 0; }