整体二分（模板二）动态区间第K大

  1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0);
  2 #include <cstdio>//sprintf islower isupper
  3 #include <cstdlib>//malloc  exit strcat itoa system("cls")
  4 #include <iostream>//pair
  5 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\Input.txt","r",stdin);
  6 #include <bitset>
  7 //#include <map>
  8 //#include<unordered_map>
  9 #include <vector>
 10 #include <stack>
 11 #include <set>
 12 #include <string.h>//strstr substr strcat
 13 #include <string>
 14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
 15 #include <cmath>
 16 #include <deque>
 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
 18 #include <vector>//emplace_back
 19 //#include <math.h>
 20 #include <cassert>
 21 #include <iomanip>
 22 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
 23 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
 24 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
 25 //******************
 26 clock_t __START,__END;
 27 double __TOTALTIME;
 28 void _MS(){__START=clock();}
 29 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
 30 //***********************
 31 #define rint register int
 32 #define fo(a,b,c) for(rint a=b;a<=c;++a)
 33 #define fr(a,b,c) for(rint a=b;a>=c;--a)
 34 #define mem(a,b) memset(a,b,sizeof(a))
 35 #define pr printf
 36 #define sc scanf
 37 #define ls rt<<1
 38 #define rs rt<<1|1
 39 typedef pair<int,int> PII;
 40 typedef vector<int> VI;
 41 typedef long long ll;
 42 const double E=2.718281828;
 43 const double PI=acos(-1.0);//
 44 const ll INF=(1LL<<60);
 45 const int inf=(1<<30);
 46 const double ESP=1e-9;
 47 const int mod=(int)1e9+7;
 48 const int N=(int)4e6+10;
 49 
 50 int ans[N],a[N];
 51 int c[N],maxn;
 52 int lowbit(int n){return n&(-n);}
 53 void add(int i,int t)
 54 {
 55     while(i<=maxn)
 56     {
 57         c[i]+=t;
 58         i+=lowbit(i);
 59     }
 60 }
 61 int Sum(int i)
 62 {
 63     int sum=0;
 64     while(i)
 65     {
 66         sum+=c[i];
 67         i-=lowbit(i);
 68     }
 69     return sum;
 70 }
 71 struct node
 72 {
 73     int l,r,k,id,currk,op;
 74 }q[N],temp[N];
 75 void Init()
 76 {
 77     for(int i=0;i<=N-3;++i)
 78         c[i]=0;
 79 }
 80 
 81 void solve(int l,int r,int ql,int qr)
 82 {
 83     if(ql>qr)return;
 84     int L=ql,R=qr,mid=l+r>>1;
 85 
 86     if(l==r)
 87     {
 88         for(int i=ql;i<=qr;++i)
 89             if(q[i].op==3)
 90                 ans[q[i].id]=l;//答案
 91         return;
 92     }
 93     for(int i=ql;i<=qr;++i)
 94     {
 95         if(q[i].op!=3)temp[q[i].r<=mid?L++:R--]=q[i];//分位置
 96         
 97         if(q[i].op==1&&q[i].r<=mid)add(q[i].l,1);//q.l才是要加的位置,q.r是值
 98         else if(q[i].op==2&&q[i].r<=mid)add(q[i].l,-1);
 99         else if(q[i].op==3)
100         {
101             int sum=Sum(q[i].r)-Sum(q[i].l-1);//二分是保证按时间顺序来的,直接计算就行
102             if(q[i].currk+sum>=q[i].k)temp[L++]=q[i];//分位置
103             else q[i].currk+=sum,temp[R--]=q[i];
104         }
105     }
106     for(int i=ql;i<=qr;++i)//清空-树状数组
107     {
108         if(q[i].op==1&&q[i].r<=mid)add(q[i].l,-1);
109         else if(q[i].op==2&&q[i].r<=mid)add(q[i].l,1);
110     }
111     //还原
112     for(int i=ql;i<=L-1;++i)
113         q[i]=temp[i];
114     for(int i=qr,j=L;j<=qr;--i,++j)
115         q[j]=temp[i];
116 
117     //递归下去
118     if(L>ql)solve(l,mid,ql,L-1);
119     if(R<qr)solve(mid+1,r,R+1,qr);
120 }
121 
122 int main()
123 {
124     int n,m;
125     while(~sc("%d",&n))
126     {
127     //    Init();
128         maxn=n;
129         for(int i=1;i<=n;++i)
130             sc("%d",&q[i].r),q[i].l=i,a[i]=q[i].r,q[i].op=1;
131         sc("%d",&m);
132         int ask=0;
133         for(int i=1;i<=m;++i)
134         {
135             int J;
136             sc("%d",&J);
137             if(J==1)
138             {
139                 int pos,x;
140                 sc("%d%d",&pos,&x);
141                 ++n;
142                 q[n].l=pos,q[n].r=a[pos],q[n].op=2;//del
143                 ++n,a[pos]=x;
144                 q[n].l=pos,q[n].r=a[pos],q[n].op=1;//add
145             }
146             else
147                 ++n,sc("%d%d%d",&q[n].l,&q[n].r,&q[n].k),q[n].currk=0,q[n].id=++ask,q[n].op=3;//query
148         }
149         solve(0,inf,1,n);
150         for(int i=1;i<=ask;++i)
151             pr("%d\n",ans[i]);
152     }
153     return 0;
154 }
155 
156 /**************************************************************************************/