HDU-4417-Super Mario(主席树解法）

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.

InputThe 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.)OutputFor 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

求一个区间比小于等于K的个数 我们就可以去查询第R个版本的线段树-第L-1个版本的线段树的数量
代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<cmath>

const int maxn=1e5+5;
typedef long long ll;
using namespace std;
struct node
{
    int l,r;
    int sum;
}tree[maxn*20];
int cnt,root[maxn];
int a[maxn];
vector<int>v;
int getid(int x)
{
    return lower_bound(v.begin(),v.end(),x)-v.begin()+1;
}
void build(int &u,int l,int r)
{
    u=++cnt;
    tree[u].sum=0;
    if(l==r)return ;
    int mid=(l+r)/2;
    build(tree[u].l,l,mid);
    build(tree[u].r,mid+1,r);
}
void update(int l,int r,int pre,int &now,int p)
{
    tree[++cnt]=tree[pre];
    now=cnt;
    tree[now].sum++; 
    if(l==r)
    {
        return ;
    }
    int mid=(l+r)>>1;
    if(p<=mid)
    {
        update(l,mid,tree[pre].l,tree[now].l,p);
    } 
    else
    {
        update(mid+1,r,tree[pre].r,tree[now].r,p);
    }
}
int query(int l,int r,int L,int R,int k)
{
  if(l==r)
  {
      return tree[R].sum-tree[L].sum;
  }
  int mid=(l+r)>>1;
  if(k<=mid)
  {
      return query(l,mid,tree[L].l,tree[R].l,k);
  }
  else
  {
      ll ans=tree[tree[R].l].sum-tree[tree[L].l].sum;
    ans+=query(mid+1,r,tree[L].r,tree[R].r,k);
    return ans;
  }
} 
int main()
{ 
    int T;
    cin>>T;
    int cc=1;
    while(T--)
    {
      int n,m;
      scanf("%d%d",&n,&m);
      cnt=0;
      for(int t=1;t<=n;t++)
      {
          v.clear();
      }
      for(int t=1;t<=n;t++)
      {
          scanf("%d",&a[t]);
          v.push_back(a[t]);
      }    
      
      sort(v.begin(),v.end());
      v.erase(unique(v.begin(),v.end()),v.end());
      build(root[0],1,n);
      for(int t=1;t<=n;t++)
      {
          update(1,n,root[t-1],root[t],getid(a[t]));
      }
      int x,y,k;
      printf("Case %d:\n",cc++);
      while(m--)
      {
          scanf("%d%d%d",&x,&y,&k);
          k=upper_bound(v.begin(),v.end(),k)-v.begin();
          if(k==0)
          {
              puts("0");
              continue;
        }
          x++;
          y++;
          printf("%d\n",query(1,n,root[x-1],root[y],k));
      }
    }
    return 0;
}