HDU 4417 Super Mario（主席树求区间内的区间查询+离散化）

Posted 2020-08-13 Blackops

Super Mario

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5101    Accepted Submission(s): 2339

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

 

 

题目链接：HDU 4417

题意：给你n个数组成的序列求[L,R]中小于等于H的数有多少个，序列下标从0开始

本来对于一般的线段树就直接求和query(1,1,H)即可，但是这样子显然默认的根节点为1即对整颗线段树进行查询，而题目中给的是固定的一段区间，并非朴素线段树默认的的1，N。然后就可以用到主席树了，主席树可以有很多种的区间查询求区间第K大只是其中一种而已吧，这题就是求区间内的区间求和，好像听起来很别扭，就是对一个固定区间内部进行值域求和，本题就是进行区间查询（求和），过程和普通线段树非常相似，在注释中对比了一下普通线段树的姿势……

由于H可能过大，因此要离散化再用lowerbound把H等效为某一个离散化后的高度

代码：

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define MID(x,y) ((x+y)>>1)

const int N=1e5+7;
struct seg
{
    int lson,rson;
    int cnt;
};
seg T[N*20];
int root[N],arr[N],tot;
vector<int>vec;

void init()
{
    CLR(root,0);
    tot=0;
    vec.clear();
}
void update(int &cur,const int &ori,const int &l,const int &r,const int &val)
{
    cur=++tot;
    T[cur]=T[ori];
    ++T[cur].cnt;
    if(l==r)
        return ;
    int mid=MID(l,r);
    if(val<=mid)
        update(T[cur].lson,T[ori].lson,l,mid,val);
    else
        update(T[cur].rson,T[ori].rson,mid+1,r,val);
}
int query(const int &S,const int &E,const int &l,const int &r,const int &x,const int &y)
{
    if(x<=l&&r<=y)//l<=T[k].l&&T[k].r<=r
        return T[E].cnt-T[S].cnt;//return T[k].cnt;
    int mid=MID(l,r);
    if(y<=mid)//r<=T[k].mid
        return query(T[S].lson,T[E].lson,l,mid,x,y);
    else if(x>mid)//l>T[k].mid
        return query(T[S].rson,T[E].rson,mid+1,r,x,y);
    else
        return query(T[S].lson,T[E].lson,l,mid, x,mid)+query(T[S].rson,T[E].rson,mid+1,r, mid+1,y);
}
int main(void)
{
    int T,n,m,i,l,r,h;
    scanf("%d",&T);
    for (int q=1; q<=T; ++q)
    {
        init();
        scanf("%d%d",&n,&m);
        for (i=1; i<=n; ++i)
        {
            scanf("%d",&arr[i]);
            vec.push_back(arr[i]);
        }
        sort(vec.begin(),vec.end());
        vec.erase(unique(vec.begin(),vec.end()),vec.end());
        int R=vec.size();
        for (i=1; i<=n; ++i)
        {
            arr[i]=lower_bound(vec.begin(),vec.end(),arr[i])-vec.begin()+1;
            update(root[i],root[i-1],1,R,arr[i]);
        }
        printf("Case %d:\n",q);
        for (i=0; i<m; ++i)
        {
            scanf("%d%d%d",&l,&r,&h);
            ++l;
            ++r;
            int indx=lower_bound(vec.begin(),vec.end(),h)-vec.begin()+1;
            //由于H可能过大因此只能找一个最接近的值indx来等效代替
            if(vec[indx-1]!=h)
                --indx;
            printf("%d\n",indx? query(root[l-1],root[r],1,R,1,indx) : 0);
        }
    }
    return 0;
}