BZOJ4552 HEOI/TJOI2016 排序 线段树、二分答案

题目传送门：https://www.lydsy.com/JudgeOnline/problem.php?id=4552

题意：给出一个$1$到$N$的全排列，对其进行$M$次排序，每次排序将区间$[l,r]$从小到大或从大到小排序，求排序完后位置$q$上的数字。$N,M \leq 10^5$，时限$6s$

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名字挂羊头卖狗肉……

暴力排序$O(NMlogN)$神级常数才能过，故考虑在排序上降低复杂度

考虑二分答案，将小于等于当前答案的设为$0$，大于当前答案的设为$1$，这样修改就变成$0,1$排序，可以使用线段树区间覆盖实现

时间复杂度$O(Mlog^2N)$

 #include<bits/stdc++.h>
 #define MAXN 100005
 using namespace std;
 inline int read(){
     ;
     char c = getchar();
     while(!isdigit(c))
         c = getchar();
     while(isdigit(c)){
         a = (a << ) + (a << ) + (c ^ ');
         c = getchar();
     }
     return a;
 }

 struct node{
     int l , r , mark , num1;
 }Tree[MAXN << ];
 ] , M , askNum , mid;

 inline void pushup(int dir){
     Tree[dir].num1 = Tree[dir << ].num1 + Tree[dir <<  | ].num1;
 }

 inline void pushdown(int dir){
     )
         Tree[dir << ].mark = Tree[dir <<  | ].mark = Tree[dir << ].num1 = Tree[dir <<  | ].num1 = ;
     else
         ){
             Tree[dir << ].mark = Tree[dir <<  | ].mark = ;
             Tree[dir << ].num1 = Tree[dir << ].r - Tree[dir << ].l + ;
             Tree[dir <<  | ].num1 = Tree[dir <<  | ].r - Tree[dir <<  | ].l + ;
         }
     Tree[dir].mark = -;
 }

 void init(int l , int r , int dir){
     Tree[dir].l = l;
     Tree[dir].r = r;
     Tree[dir].mark = -;
     if(l == r)
         Tree[dir].num1 = num[l] > mid;
     else{
         init(l , l + r >>  , dir << );
         init((l + r >> ) +  , r , dir <<  | );
         pushup(dir);
     }
 }

 void change(int l , int r , int dir , int mark){
     if(Tree[dir].l >= l && Tree[dir].r <= r){
         Tree[dir].mark = mark;
         Tree[dir].num1 = mark * (Tree[dir].r - Tree[dir].l + );
         return;
     }
     pushdown(dir);
     )
         change(l , r , dir <<  , mark);
     )
         change(l , r , dir <<  |  , mark);
     pushup(dir);
 }

 int ask(int l , int r , int dir){
     if(Tree[dir].l >= l && Tree[dir].r <= r)
         return Tree[dir].num1;
     ;
     pushdown(dir);
     )
         sum += ask(l , r , dir << );
     )
         sum += ask(l , r , dir <<  | );
     return sum;
 }

 inline bool check(){
     init( , N , );
      ; i <= M ; i++){
         ] , sortNum[i][] , );
         ]){
             change(sortNum[i][] , sortNum[i][] + t -  ,  , );
             ] + t <= sortNum[i][])
                 change(sortNum[i][] + t , sortNum[i][] ,  , );
         }
         else{
             change(sortNum[i][] - t + ,  sortNum[i][] ,  , );
             ] - t >= sortNum[i][])
                 change(sortNum[i][] , sortNum[i][] - t ,  , );
         }
     }
     );
 }
 int main(){
     N = read();
     M = read();
      ; i <= N ; i++)
         num[i] = read();
      ; i <= M ; i++){
         sortNum[i][] = read();
         sortNum[i][] = read();
         sortNum[i][] = read();
     }
     askNum = read();
      , r = N;
     while(l < r){
         mid = l + r >> ;
         if(check())
             r = mid;
         else
             l = mid + ;
     }
     cout << l;
     ;
 }