数据结构(线段树)：HDU 5649 DZY Loves Sorting

DZY Loves Sorting

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 294    Accepted Submission(s): 77

Problem Description

　　DZY has a sequence a[1..n]. It is a permutation of integers 1∼n.
　　Now he wants to perform two types of operations:
　　　　0 l r: Sort a[l..r] in increasing order.
　　　　1 l r: Sort a[l..r] in decreasing order.
　　After doing all the operations, he will tell you a position k, and ask you the value of a[k].

Input

　　First line contains t, denoting the number of testcases.
　　t testcases follow. For each testcase:
　　First line contains n,m. m is the number of operations.
　　Second line contains n space-separated integers a[1],a[2],⋯,a[n], the initial sequence. We ensure that it is a permutation of 1∼n.
　　Then m lines follow. In each line there are three integers opt,l,r to indicate an operation.
　　Last line contains k.
　　(1≤t≤50,1≤n,m≤100000,1≤k≤n,1≤l≤r≤n,opt∈{0,1}. Sum of n in all testcases does not exceed 150000. Sum of m in all testcases does not exceed 150000)

Output

　　For each testcase, output one line - the value of a[k] after performing all m operations.

Sample Input

　　1

　　6 3

　　1 6 2 5 3 4

　　0 1 4

　　1 3 6

　　0 2 4

　　3

Sample Output

　　5

Hint

1 6 2 5 3 4 -> [1 2 5 6] 3 4 -> 1 2 [6 5 4 3] -> 1 [2 5 6] 4 3. At last a[3]=5.

 

　　好题，考虑二分的话，维护的只是相对大小，题目就好做了。

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 using namespace std;
 const int maxn=;
 int n,m,k;
 int tr[maxn<<],mark[maxn<<];
 int L[maxn],R[maxn],X[maxn],a[maxn];

 void Make_same(int x,int l,int r,int d){
     tr[x]=(r-l+)*d;
     mark[x]=d;
 }

 void Push_down(int x,int l,int r){
     if(mark[x]!=-){
         int mid=(l+r)>>;
         Make_same(x<<,l,mid,mark[x]);
         Make_same(x<<|,mid+,r,mark[x]);
         mark[x]=-;
     }
 }

 void Build(int x,int l,int r,int g){
     mark[x]=-;
     if(l==r){
         tr[x]=a[l]<=g?:;
         return;
     }
     int mid=(l+r)>>;
     Build(x<<,l,mid,g);
     Build(x<<|,mid+,r,g);
     tr[x]=tr[x<<]+tr[x<<|];
 }

 int Query(int x,int l,int r,int a,int b){
     Push_down(x,l,r);
     if(l>=a&&r<=b)return tr[x];
     int mid=(l+r)>>,ret=;
     if(mid>=a)ret=Query(x<<,l,mid,a,b);
     if(mid<b)ret+=Query(x<<|,mid+,r,a,b);
     return ret;
 }

 void Mark(int x,int l,int r,int a,int b,int d){
     if(a>b)return;
     if(l>=a&&r<=b){
         Make_same(x,l,r,d);
         return;
     }
     Push_down(x,l,r);
     int mid=(l+r)>>;
     if(mid>=a)Mark(x<<,l,mid,a,b,d);
     if(mid<b)Mark(x<<|,mid+,r,a,b,d);
     tr[x]=tr[x<<]+tr[x<<|];
 }

 bool Check(){
     for(int i=;i<=m;i++){
         int sum=Query(,,n,L[i],R[i]);
         if(X[i]){
             Mark(,,n,L[i],L[i]+sum-,);
             Mark(,,n,L[i]+sum,R[i],);
         }
         else{
             sum=R[i]-L[i]+-sum;
             Mark(,,n,L[i],L[i]+sum-,);
             Mark(,,n,L[i]+sum,R[i],);
         }
     }
     return Query(,,n,k,k);
 }

 void Solve(){
     int lo=,hi=n;
     while(lo<=hi){
         int mid=(lo+hi)>>;
         Build(,,n,mid);
         if(Check())lo=mid+;
         else hi=mid-;
     }
     printf("%d\n",lo);
     return;
 }
 int main(){
     int T;
     scanf("%d",&T);
     while(T--){
         scanf("%d%d",&n,&m);
         for(int i=;i<=n;i++)scanf("%d",&a[i]);
         for(int i=;i<=m;i++)scanf("%d%d%d",&X[i],&L[i],&R[i]);
         scanf("%d",&k);
         Solve();
     }
     return ;
 }