HDU 4417 Super Mario（主席树）

Posted 2020-10-31 hinata_hajime

Super Mario

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8365    Accepted Submission(s): 3540

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

 

 

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题意是求 区间[L,R]的小于等于h的数的个数

 

主席树的模板题

 

代码如下：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<map>
#include<stack>
#include<vector>
using namespace std;
const int MAXN=1e5+10;
int a[MAXN],root[MAXN];
int t,n,m,l,r,x;
vector<int>v;
int get_id(int x){return lower_bound(v.begin(),v.end(),x)-v.begin()+1;}  //因为每个数的范围很大，需要进行离散化处理

struct node
{
  int l;
  int r;
  int h;
}qes[MAXN];

struct Node
{
   int l;
   int r;
   int sum;
}T[MAXN*40];
int cnt;
int ans;
void Update(int l,int r,int &x,int y,int pos)
{
   T[++cnt]=T[y],T[cnt].sum++,x=cnt;
   if(l==r)return;
   int mid=(l+r)>>1;
   if(pos<=mid)Update(l,mid,T[x].l,T[y].l,pos);
   else Update(mid+1,r,T[x].r,T[y].r,pos);
}

void query(int l,int r,int x,int y,int k)
{

    int mid=(l+r)>>1;
    if(mid==k)// 如果小于等于k的数刚好全部在左边区域，加上它的权值，直接结束
    {
    ans+=T[T[y].l].sum-T[T[x].l].sum;
    return;
    }
    else if(k<mid) //左边存在部分小于等于k的数，查询左边区域
    query(l,mid,T[x].l,T[y].l,k);
    else//右边存在部分小于k的数，则加上左边的权值，并查询右边区域。
    {
      ans+=T[T[y].l].sum-T[T[x].l].sum;
      query(mid+1,r,T[x].r,T[y].r,k);
    }
}

int main()
{
    scanf("%d",&t);
    int Case=0;
    while(t--)
    {
     Case++;
     cnt=0;
     v.clear();
     scanf("%d%d",&n,&m);
     for(int i=1;i<=n;i++)
     {
      scanf("%d",&a[i]);
      v.push_back(a[i]);
     }

      for(int i=1;i<=m;i++)
      {
         scanf("%d%d%d",&qes[i].l,&qes[i].r,&qes[i].h);
          v.push_back(qes[i].h);
      }
       sort(v.begin(),v.end());
       v.erase(unique(v.begin(),v.end()),v.end());



      int sz=v.size();
      for(int i=1;i<=n;i++)
      Update(1,sz,root[i],root[i-1],get_id(a[i]));

      printf("Case %d:\n",Case);
      for(int i=1;i<=m;i++)
      {
         ans=0;
         query(1,sz,root[qes[i].l],root[qes[i].r+1],get_id(qes[i].h));
         printf("%d\n",ans);
      }

    }
    return 0;
}