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=100010;
int n,m,k;
int tr[maxn<<2],mark[maxn<<2];
int L[maxn],R[maxn],X[maxn],a[maxn];

void Make_same(int x,int l,int r,int d){
    tr[x]=(r-l+1)*d;
    mark[x]=d;
}

void Push_down(int x,int l,int r){
    if(mark[x]!=-1){
        int mid=(l+r)>>1;
        Make_same(x<<1,l,mid,mark[x]);
        Make_same(x<<1|1,mid+1,r,mark[x]);
        mark[x]=-1;
    }
}

void Build(int x,int l,int r,int g){
    mark[x]=-1;
    if(l==r){
        tr[x]=a[l]<=g?0:1;
        return;
    }
    int mid=(l+r)>>1;
    Build(x<<1,l,mid,g);
    Build(x<<1|1,mid+1,r,g);
    tr[x]=tr[x<<1]+tr[x<<1|1];
}

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)>>1,ret=0;
    if(mid>=a)ret=Query(x<<1,l,mid,a,b);
    if(mid<b)ret+=Query(x<<1|1,mid+1,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)>>1;
    if(mid>=a)Mark(x<<1,l,mid,a,b,d);
    if(mid<b)Mark(x<<1|1,mid+1,r,a,b,d);
    tr[x]=tr[x<<1]+tr[x<<1|1];    
}

bool Check(){
    for(int i=1;i<=m;i++){
        int sum=Query(1,1,n,L[i],R[i]);
        if(X[i]){
            Mark(1,1,n,L[i],L[i]+sum-1,1);
            Mark(1,1,n,L[i]+sum,R[i],0);
        }
        else{
            sum=R[i]-L[i]+1-sum;
            Mark(1,1,n,L[i],L[i]+sum-1,0);
            Mark(1,1,n,L[i]+sum,R[i],1);
        }
    }
    return Query(1,1,n,k,k);
}

void Solve(){
    int lo=1,hi=n;
    while(lo<=hi){
        int mid=(lo+hi)>>1;
        Build(1,1,n,mid);
        if(Check())lo=mid+1;
        else hi=mid-1;    
    }
    printf("%d
",lo);
    return;
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        for(int i=1;i<=m;i++)scanf("%d%d%d",&X[i],&L[i],&R[i]);
        scanf("%d",&k);
        Solve();
    }
    return 0;
}