hdu5649 DZY Loves Sorting

http://www.elijahqi.win/archives/2935
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:

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 a[3]=5 a [ 3 ] = 5 .

每次二分下答案 然后将比他大的数改成1比他小的数改成0 那么每次对一个区间排序显然就是对0，1排序 那么查询一下区间内有多少个0，1即可 然后直接线段树区间修改 然后最后询问p位置上是0 还是1 如果是1 说明答案还可以再大 一直二分即可

#include
#include
#include
#define lc (x<<1)
#define rc (x<<1|1)
#define N 100010
using namespace std;
inline char gc(){
    static char now[1<<16],*S,*T;
    if(T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0,f=1;char ch=gc();
    while(!isdigit(ch)) {if (ch=='-') f=-1;ch=gc();}
    while(isdigit(ch)) x=x*10+ch-'0',ch=gc();
    return x*f;
}
struct node{
    int s,tag;
}tree[N<<2];
inline void pushdown(int x,int l,int r,int mid){
    if (tree[x].tag==-1) return;
    if (tree[x].tag==1){tree[x].tag=-1;
        tree[lc].s=mid-l+1;tree[lc].tag=1;
        tree[rc].s=r-mid;tree[rc].tag=1;return;
    }tree[x].tag=-1;
    tree[lc].s=tree[rc].s=tree[lc].tag=tree[rc].tag=0;
}
inline void update(int x){tree[x].s=tree[lc].s+tree[rc].s;}
inline void modify(int x,int l,int r,int l1,int r1,int v){
    if (l1>r1) return;
    if (l1<=l&&r1>=r){tree[x].tag=v;tree[x].s=v*(r-l+1);return;}
    int mid=l+r>>1;pushdown(x,l,r,mid);
    if (l1<=mid) modify(lc,l,mid,l1,r1,v);
    if (r1>mid) modify(rc,mid+1,r,l1,r1,v);update(x);
}int ave,a[N];
inline void build(int x,int l,int r){
    if(l==r) {tree[x].s=(a[l]>=ave);return;}int mid=l+r>>1;
    tree[x].tag=-1;build(lc,l,mid);build(rc,mid+1,r);update(x);
}
inline int query(int x,int l,int r,int l1,int r1){
    if (l1<=l&&r1>=r) return tree[x].s;
    int mid=l+r>>1;pushdown(x,l,r,mid);int tmp=0;
    if (l1<=mid) tmp+=query(lc,l,mid,l1,r1);
    if (r1>mid) tmp+=query(rc,mid+1,r,l1,r1);return tmp;
}
inline void print(int x,int l,int r){
    if(l==r) {printf("%d %d %d\n",l,r,tree[x].s);return;}
    int mid=l+r>>1;pushdown(x,l,r,mid);
    print(lc,l,mid);printf("%d %d %d\n",l,r,tree[x].s);
    print(rc,mid+1,r);
}
int n,m,l[N],r[N],op[N],p;
inline bool check(int md){
    ave=md;build(1,1,n);//printf("%d\n",md);print(1,1,n);puts("---");
    for (int i=1;i<=m;++i){static int s,s1;
        if(op[i]==0){
            s=query(1,1,n,l[i],r[i]);s1=r[i]-l[i]+1-s;
            modify(1,1,n,l[i],l[i]+s1-1,0);
            modify(1,1,n,r[i]-s+1,r[i],1);//print(1,1,n);puts("---");
        }else{
            s=query(1,1,n,l[i],r[i]);s1=r[i]-l[i]+1-s;
            modify(1,1,n,l[i],l[i]+s-1,1);//print(1,1,n);puts("---");
            modify(1,1,n,r[i]-s1+1,r[i],0);//print(1,1,n);puts("---");
        }
    }return query(1,1,n,p,p);
}
int main(){
    //freopen("bzoj4552.in","r",stdin);
    int T=read();
    while(T--){
        n=read();m=read();int L=1,R=n;
        for (int i=1;i<=n;++i) a[i]=read();
        for (int i=1;i<=m;++i) op[i]=read(),l[i]=read(),r[i]=read();p=read();
        while(L<=R){static int mid;
            mid=L+R>>1;
            if (check(mid)) L=mid+1;else R=mid-1;
        }printf("%d\n",R);
    }
    return 0;
}