bzoj4552tjoi2016&heoi2016 Sort Segment Tree

DescriptionIn 2016, her sister liked the number sequence. So he often studies some weird questions about sequences, and now he's studying a problem that needs you to help him. This puzzle is like this: give a 1 to n full arrangement, now the full permutation of the sequence of M-order, sorted into two kinds: 1: (0,l,r) for the interval [l,r] of the number ascending sort 2: (1,l,r) for the interval [l,r] number descending sort finally ask Q The number on the position. InputThe first behavior of the input data is two integers n and M. n indicates the length of the sequence, and M indicates the number of local sorts. 1 <= N, M <= 10^5 the second behavior n integers, representing a full array of 1 to N. Next enter M line, each line has three integer op, L, R, op 0 for ascending sort, op 1 for descending sort, L, R for sorting interval. Finally, enter an integer q,q to indicate where to ask after sorting, 1 <= q <= N. 1 <= N <= 10^5, 1 <= m <= 10^5 Output

The output data has only one row, and an integer that represents the number at the Q position after the whole part is sorted in order. Sample Input 6 3
1 6 2 5 3 4
0 1 4
1 3 6
0 2 4
3 Sample Output 5 Solution: The first two-point answer. Then just judge whether the number is legal. Set this number to x so you can count the number in this sequence to be 1, the number smaller than X is 1, and the number equal to X is 0. Finally, only the number in the Q position can be judged to be 0. To maintain the number of -1,0,1 in the sequence, it is clear that the operation can be processed with a line tree. Code:

#include <iostream> #include <cstdio> #include <cstring> #define N 100010 using namespace std;
int a[n],ans,n,l,r,q,m,v[n],p[n<<2],sum; struct Seg{int s0,s1,s2;}
t[n<<2]; struct Use{int k,l,r;}
C[n];
  SEG Update (SEG a,seg b) {seg C;
  C.s0=a.s0+b.s0;
  C.S1=A.S1+B.S1;
  C.S2=A.S2+B.S2;
return C;
  } void Build (int k,int l,int r) {int mid= (L+R) >>1;
    if (l==r) {if (v[l]==0) t[k].s0++;
    if (v[l]==1) t[k].s1++;
    if (v[l]==2) t[k].s2++; 
  Return
  } build (K<<1,l,mid);
  Build (K<<1|1,mid+1,r); 
T[k]=update (t[k<<1],t[k<<1|1]);
   } void Paint (int k,int l,int R,int v) {if (v==0) {t[k].s0=r-l+1;t[k].s1=t[k].s2=0;}
   if (v==1) {t[k].s1=r-l+1;t[k].s0=t[k].s2=0;}
   if (v==2) {t[k].s2=r-l+1;t[k].s0=t[k].s1=0;} 
P[k]=v;
  } void pushdown (int k,int l,int r) {int mid= (L+R) >>1;
  Paint (K<<1,l,mid,p[k]);
  Paint (K<<1|1,mid+1,r,p[k]);
P[k]=-1; } SEG Query (int k,int l,int r,int ll,int rr) {int mid= (L+R) &Gt;>1;
  if (LL==L&AMP;&AMP;R==RR) {return t[k];}
  if (p[k]!=-1) pushdown (k,l,r);
  if (rr<=mid) return query (K&LT;&LT;1,L,MID,LL,RR);
  else if (MID&LT;LL) return query (K&LT;&LT;1|1,MID+1,R,LL,RR); 
else return update (query (K&LT;&LT;1,L,MID,LL,MID), query (K&LT;&LT;1|1,MID+1,R,MID+1,RR));
  } void Change (int k,int l,int r,int ll,int rr,int v) {int mid= (L+R) >>1; 
  if (LL&LT;=L&AMP;&AMP;R&LT;=RR) {paint (k,l,r,v); return;}
  if (p[k]!=-1) pushdown (k,l,r);
  if (ll<=mid) change (K&LT;&LT;1,L,MID,LL,RR,V);
  if (MID&LT;RR) change (K&LT;&LT;1|1,MID+1,R,LL,RR,V); 
T[k]=update (t[k<<1],t[k<<1|1]);
    } int judge (int x) {for (int i=1;i<=n;i++) if (a[i]<x) v[i]=0;
    else if (a[i]==x) v[i]=1;
  else v[i]=2;
  for (int i=1;i<= (N&LT;&LT;2); i++) T[i].s0=t[i].s1=t[i].s2=0,p[i]=-1;
  Build (1,1,n);
    for (int i=1;i<=m;i++) {seg tt=query (1,1,N,C[I].L,C[I].R); 
    cout<<tt.s0<< ' <<tt.s1<< ' <<tt.s2<<endl; If(c[i].k==0)
      {int a=c[i].l+tt.s0-1,b=a+tt.s1;
      if (TT.S0) change (1,1,n,c[i].l,a,0);
      cout<<a+1<< ' <<b<<endl;
      if (TT.S1) change (1,1,n,a+1,b,1); 
    if (TT.S2) change (1,1,n,b+1,c[i].r,2);
      } else{int a=c[i].l+tt.s2-1,b=a+tt.s1;
      if (TT.S2) change (1,1,n,c[i].l,a,2);
      if (TT.S1) change (1,1,n,a+1,b,1);    
    if (TT.S0) change (1,1,n,b+1,c[i].r,0);
  }} seg Tt=query (1,1,N,Q,Q);
  if (TT.S0) return 0;
  if (TT.S1) return 1; 
if (TT.S2) return 2;
   } int main () {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", &AMP;C[I].K,&AMP;C[I].L,&AMP;C[I].R);
   scanf ("%d", &q); l=1;r=n;
   Cout<<judge (5) <<endl;
     while (l<=r) {//cout<<l<< ' <<r<<endl;
     int mid= (L+R) >>1; 
     int T=judge (mid);
     if (t==2) l=mid+1;
     else if (t==1) {ans=mid;break;}
   else r=mid-1; } cout<<ans<<endl;  }