HDU - 5649 线段树+区间二分答案 （好题）

DZY has a sequence a[1…n]. It is a permutation of integers 1∼n.

Now he wants to perform two types of operations:

0lr: Sort a[l…r] in increasing order.

1lr: 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 .

这次题目之前我们数据结构专题出过，所以比赛时候就直接交了；
考虑朴素算法：
每次对区间进行 sort,假设平均复杂度 NlogN
算上操作次数t. TLE是肯定的；
思路：
考虑用线段树进行维护，我们可以将大于 mid的设为1，小于mid 设为0 ，那么升序的时候将后面>mid设为1；降序则将前面设为1.每次二分答案，如果查询 [k,k]区间==1，说明在左区间；反之在右区间；

#include<bits/stdc++.h>
 using namespace std;
const int maxn = 2e5+7;
int lazy[maxn<<2];
int s[maxn<<2];
int a[maxn];
int op[maxn];
int l[maxn];
int r[maxn];
int n,k;
int tot;

void pushDown( int rt , int L , int R ){
    if( lazy[rt] != -1 ){
        int mid = ( L + R ) >> 1 ;
        s[rt<<1] = ( mid - L + 1 ) * lazy[rt] ;
        s[rt<<1|1] = ( R - mid ) * lazy[rt];
        lazy[rt<<1] = lazy[rt];
        lazy[rt<<1|1] = lazy[rt];
        lazy[rt] = -1;
    }
    return ;
}


void pushUp( int rt ){
    s[rt] = s[rt<<1] + s[rt<<1|1];
    return ;
}

void build( int l , int r , int rt , int v ){
    lazy[rt] = -1;
    if( l == r ){
        s[rt] =  ( a[l] > v ) ;
        return ; 
    }
    int mid = ( l + r ) >> 1 ;
    build( l , mid , rt << 1 , v );
    build( mid + 1 , r , rt << 1 | 1 , v);
    pushUp(rt);
}

void update( int L , int R , int v , int l , int r , int rt ){
    if( L <= l && R >= r ){
        s[rt] = v * ( r - l + 1 );
        lazy[rt] = v;
        return;
    }
    pushDown(rt,l,r);
    int mid = ( l + r ) >> 1;
    if( L <= mid ) update( L , R , v , l , mid , rt << 1 );
    if( R > mid ) update( L , R , v , mid + 1 , r , rt << 1 | 1);
    pushUp(rt);
}

int query( int L , int R , int l , int r , int rt ){
    if( L <= l && R >= r ){
        return s[rt];
    }
    pushDown(rt,l,r);
    int ans = 0 ;
    int mid = ( l + r ) >> 1;
    if( L <= mid ) ans += query( L , R , l , mid, rt << 1 );
    if( R > mid )  ans += query( L , R , mid + 1, r , rt << 1 | 1 );
    return ans ;
}

int main(){
	int t;
	cin>>t;
	while(t--){
	memset(s,0,sizeof(s));
	memset(lazy,0,sizeof(lazy));
	memset(a,0,sizeof(a));
	memset(op,0,sizeof(op));
	memset(l,0,sizeof(l));
	memset(r,0,sizeof(r));
    scanf("%d",&n);
	int m ;
    scanf("%d",&m);
    for( int i = 1 ; i <= n ; i++ ){
        scanf("%d",&a[i]);
    }
    
    for( int i = 0 ; i < m ; i++ ){
        scanf("%d%d%d",&op[i],&l[i],&r[i]);     
    }
    cin>>k;
    int li = 1 , ri = n ; 
    while( li < ri ){
        int mid = ( li + ri ) >> 1 ;
        build( 1 , n , 1 , mid );
        for( int i = 0 ; i < m ; i++ ){
            int L = l[i];
            int R = r[i];
            int c = query( L , R , 1 , n , 1 );
            update( L , R , 0 , 1 , n , 1 );
            if( op[i] ){  
                if( L <= L + c - 1 ){
                    update( L , L + c - 1 , 1 , 1, n , 1 );
                } 
            }else{
                if( R - c + 1 <= R ){
                    update( R - c + 1 , R , 1 , 1 , n , 1 );
                }
            }
        }
        if( query( k , k , 1 , n , 1 ) ){
            li = mid + 1 ;
        }else ri = mid  ;
    }
    printf("%d\n",li);
}
    return 0 ;
}