2012 ACM/ICPC Asia Regional Hangzhou Online ，HDU4417-Super Mario

Problem Description

Mario is world-famous plumber. His “burly” figure and amazing jumping ability reminded in our memory. Now the poor princess is in trouble again and Mario needs to save his lover. We regard the road to the boss’s castle as a line (the length is n), on every integer point i there is a brick on height hi. Now the question is how many bricks in [L, R] Mario can hit if the maximal height he can jump is H.

 

Input

The first line follows an integer T, the number of test data.
For each test data:
The first line contains two integers n, m (1 <= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the number of queries.
Next line contains n integers, the height of each brick, the range is [0, 1000000000].
Next m lines, each line contains three integers L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.)

 

Output

For each case, output "Case X: " (X is the case number starting from 1) followed by m lines, each line contains an integer. The ith integer is the number of bricks Mario can hit for the ith query.

 

Sample Input

    
    
    
    

     
     
     
     1
10 10
0 5 2 7 5 4 3 8 7 7 
2 8 6
3 5 0
1 3 1
1 9 4
0 1 0
3 5 5
5 5 1
4 6 3
1 5 7
5 7 3
    
    
    
    

 

Sample Output

    
    
    
    

     
     
     
     Case 1:
4
0
0
3
1
2
0
1
5
1
    
    
    
    

 

Source

2012 ACM/ICPC Asia Regional Hangzhou Online

 

题意就是给你一个区间，一个数，问你这个区间里比这个数小的数有多少个，思路是划分树+二分枚举第k大的数

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>

using namespace std;

const int maxn=100010;

struct node
{
    int  num[maxn];
    int  val[maxn];
}tree[30];

int sorted[maxn];

void build(int ind,int l,int r)
{
    if(l == r)
      return ;
    int mid=(l+r)>>1;
    int isame=mid-l+1;
    for(int i=l;i<=r;i++)
      if(sorted[mid] > tree[ind].val[i])
        isame--;
    int ln=l;
    int rn=mid+1;
    for(int i = l;i<=r;i++)
    {
        if(i == l)
          tree[ind].num[i]=0;
          else
            tree[ind].num[i]=tree[ind].num[i-1];
        if(tree[ind].val[i] < sorted[mid])
        {
            tree[ind].num[i]++;
            tree[ind+1].val[ln++]=tree[ind].val[i];
        }
        else if(tree[ind].val[i] > sorted[mid])
           tree[ind+1].val[rn++]=tree[ind].val[i];
        else
        {
            if(isame)
            {
                isame--;
                tree[ind].num[i]++;
                tree[ind+1].val[ln++]=tree[ind].val[i];
            }
            else
                tree[ind+1].val[rn++]=tree[ind].val[i];
        }
    }
    build(ind+1,l,mid);
    build(ind+1,mid+1,r);
}

int query(int ind,int l,int r,int a,int b,int k)
{
    if(a == b)
      return tree[ind].val[a];
    int s0,s1;
    int mid=(l+r)>>1;
    if(a == l)
    {
        s0=0;
        s1=tree[ind].num[b];
    }
    else
    {
        s0=tree[ind].num[a-1];
        s1=tree[ind].num[b]-s0;
    }
    if(s1 >= k)
    {
        a=l+s0;
        b=l+s0+s1-1;
        return query(ind+1,l,mid,a,b,k);
    }
    else
    {
        a=mid-l+1+a-s0;
        b=mid-l+1+b-s0-s1;
        return query(ind+1,mid+1,r,a,b,k-s1);
    }
}

int main()
{
    int t;
    scanf("%d",&t);
        int p=1;
        while(t--)
        {
            int n, m, x, y;
            int k;
            scanf("%d%d",&n,&m);
            for(int i=1;i<=n;i++)
            {
                scanf("%d",&sorted[i]);
                tree[0].val[i]=sorted[i];
            }
            sort(sorted+1,sorted+n+1);
            build(0,1,n);
            printf("Case %d:\n",p++);
            while(m--)
            {
                scanf("%d%d%d",&x,&y,&k);
                x++;
                y++;
                int l=1,r=(y-x+1),mid,ans=0;
                while(l<=r)
                {
                    int mid=(l+r)>>1;
                    if(query(0,1,n,x,y,mid) <= k)//第mid大元素小于k
                    {
                         l=mid+1;
                         ans=mid;
                    }  
                    else
                       r=mid-1; 
                }
                 printf("%d\n",ans);
            }
        }
    return 0;
}