POJ 2104(K-th Number-区间第k大-主席树)-白红宇

POJ 2104(K-th Number-区间第k大-主席树)

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发布时间：2019-06-29

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K-th Number

Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions: 31790		Accepted: 9838
Case Time Limit: 2000MS

Description

You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment.

That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?"

For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.

Input

The first line of the input file contains n --- the size of the array, and m --- the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000).

The second line contains n different integer numbers not exceeding 10

⁹ by their absolute values --- the array for which the answers should be given.

The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).

Output

For each question output the answer to it --- the k-th number in sorted a[i...j] segment.

Sample Input

7 31 5 2 6 3 7 42 5 34 4 11 7 3

Sample Output

Hint

This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.

Source

, Northern Subregion

这题是主席树裸题。

但是我却狂T——

#include
    
     #include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Rep(i,n) for(int i=0;i
             
              =0;i--)#define MAXN (100000+10)#define MAXM (5000+10)int n,m,a[MAXN],a2[MAXN];struct node{ node *ch[2]; int a,siz; node(){ch[0]=ch[1]=NULL;siz=a=0;} node(node *_ch0,node *_ch1,int _a,int _siz):a(_a),siz(_siz){ch[0]=_ch0,ch[1]=_ch1;} void update() { siz=a; if (ch[0]) siz+=ch[0]->siz; if (ch[1]) siz+=ch[1]->siz; }}*null=new node(),*root[MAXN]={NULL};void make_node(node *&y,node *&x,int l,int r,int t){ if (x==NULL) x=null; y=new node(); int m=(l+r)>>1; if (l==r) { *y=*x; y->siz++;y->a++; return; } if (t<=a2[m]) { make_node(y->ch[0],x->ch[0],l,m,t); y->ch[1]=x->ch[1]; y->update(); } else { make_node(y->ch[1],x->ch[1],m+1,r,t); y->ch[0]=x->ch[0]; y->update(); }}void find(node *&x1,node *&x2,int l,int r,int k){ if (x1==NULL) x1=null; if (x2==NULL) x2=null; if (l==r) {printf("%d\n",a2[l]);return;} int m=(l+r)>>1; int ls=0; if (x2->ch[0]) ls+=x2->ch[0]->siz; if (x1->ch[0]) ls-=x1->ch[0]->siz; if (ls>=k) find(x1->ch[0],x2->ch[0],l,m,k); else find(x1->ch[1],x2->ch[1],m+1,r,k-ls);}int main(){ freopen("poj2104.in","r",stdin); null->ch[0]=null; null->ch[1]=null; scanf("%d%d",&n,&m); For(i,n) scanf("%d",&a[i]),a2[i]=a[i]; sort(a2+1,a2+1+n); int size=unique(a2+1,a2+1+n)-(a2+1); For(i,n) { make_node(root[i],root[i-1],1,size,a[i]); } For(i,m) { int l,r,k; scanf("%d%d%d",&l,&r,&k); find(root[l-1],root[r],1,size,k); } return 0;}

于是在柯老师的教导下，我发现致使我狂T的因素是不断new 结点，于是我将结点事先用数组开好。

#include
    
     #include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Rep(i,n) for(int i=0;i
             
              =0;i--)#define MAXN (200000+10)#define MAXM (200000+10)int n,m,a[MAXN],a2[MAXN];struct node{ node *ch[2]; int a,siz; node(){ch[0]=ch[1]=NULL;siz=a=0;} void update() { siz=a; if (ch[0]) siz+=ch[0]->siz; if (ch[1]) siz+=ch[1]->siz; }}*null=new node(),*root[MAXN]={NULL},q[MAXN*9];int q_s;void make_node(node *&y,node *&x,int l,int r,int t){ if (x==NULL) x=null; y=&q[++q_s]; *y=node(); int m=(l+r)>>1; if (l==r) { *y=*x; y->siz++;y->a++; return; } if (t<=a2[m]) { make_node(y->ch[0],x->ch[0],l,m,t); y->ch[1]=x->ch[1]; y->update(); } else { make_node(y->ch[1],x->ch[1],m+1,r,t); y->ch[0]=x->ch[0]; y->update(); }}void find(node *&x1,node *&x2,int l,int r,int k){ if (x1==NULL) x1=null; if (x2==NULL) x2=null; if (l==r) {printf("%d\n",a2[l]);return;} int m=(l+r)>>1; int ls=0; if (x2->ch[0]) ls+=x2->ch[0]->siz; if (x1->ch[0]) ls-=x1->ch[0]->siz; if (ls>=k) find(x1->ch[0],x2->ch[0],l,m,k); else find(x1->ch[1],x2->ch[1],m+1,r,k-ls);}int main(){ //freopen("hdu2665.in","r",stdin); null->ch[0]=null; null->ch[1]=null; q_s=0; scanf("%d%d",&n,&m); For(i,n) scanf("%d",&a[i]),a2[i]=a[i]; sort(a2+1,a2+1+n); int size=unique(a2+1,a2+1+n)-(a2+1); For(i,n) { make_node(root[i],root[i-1],1,size,a[i]); } For(i,m) { int l,r,k; scanf("%d%d%d",&l,&r,&k); find(root[l-1],root[r],1,size,k); } return 0;}

终于过了……伤不起……

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