【线段树】 hdu4325 Flowers

Flowers

http://acm.hdu.edu.cn/showproblem.php?pid=4325

Problem Description

As is known to all, the blooming time and duration varies between different kinds of flowers. Now there is a garden planted full of flowers. The gardener wants to know how many flowers will bloom in the garden in a specific time. But there are too many flowers in the garden, so he wants you to help him.

 

Input

The first line contains a single integer t (1 <= t <= 10), the number of test cases.
For each case, the first line contains two integer N and M, where N (1 <= N <= 10^5) is the number of flowers, and M (1 <= M <= 10^5) is the query times. 
In the next N lines, each line contains two integer S _i and T _i (1 <= S _i <= T _i <= 10^9), means i-th flower will be blooming at time [S _i, T _i].
In the next M lines, each line contains an integer T _i, means the time of i-th query.

 

Output

For each case, output the case number as shown and then print M lines. Each line contains an integer, meaning the number of blooming flowers.
Sample outputs are available for more details.

 

Sample Input

   
   
   
   

    
    
    
    2
1 1
5 10
4
2 3
1 4
4 8
1
4
6
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1:
0
Case #2:
1
2
1
   
   
   
   

题意：给出花开放和凋谢的时间点，询问在一个时间点上有多少花同时开放。

题解：成段更新，单点查询的线段树，就是花开谢时间要hash处理，不然必然超内存。就是这个hash处理个人觉得我做的有点麻烦了。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAX 100005
struct node
{
    int s,t;
}num[MAX];
int bloom[MAX<<2];
int add[MAX<<2];
void build(int l,int r,int idx)
{
    add[idx]=0;
    if(l==r) return;
    int mid=(l+r)>>1;
    build(l,mid,idx<<1);
    build(mid+1,r,idx<<1|1);
}
int binarySearch(int x,int n)
{
    int l=0,r=n-1,mid;
    for(;l<=r;)
    {
        mid=(l+r)>>1;
        if(x<bloom[mid])
           r=mid-1;
        else if(x==bloom[mid])
           return mid;
        else
           l=mid+1;
    }
    return l;
}
void push_down(int l,int r,int idx)
{
    if(add[idx])
    {
        add[idx<<1]+=add[idx];
        add[idx<<1|1]+=add[idx];
        add[idx]=0;
    }
}
void update(int a,int b,int l,int r,int idx)
{
    if(a<=l&&r<=b)
    {
        add[idx]++;
        return;
    }
    push_down(l,r,idx);
    int mid=(l+r)>>1;
    if(a<=mid) update(a,b,l,mid,idx<<1);
    if(b>mid)  update(a,b,mid+1,r,idx<<1|1);
}
int query(int x,int l,int r,int idx)
{
    if(l==x&&x==r)
       return add[idx];
    push_down(l,r,idx);
    int mid=(l+r)>>1;
    if(x<=mid) return query(x,l,mid,idx<<1);
    else       return query(x,mid+1,r,idx<<1|1);
}
int main()
{
    int ti,n,m,cnt,ans;
    scanf("%d",&ti);
    for(int cas=1;cas<=ti;++cas)
    {
        cnt=1;
        printf("Case #%d:\n",cas);
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;++i)
        {
            scanf("%d%d",&num[i].s,&num[i].t);
            bloom[cnt++]=num[i].s;
            bloom[cnt++]=num[i].t;
        }
        //以下就是hash处理，就是把时间点的-1和+1都补上
        bloom[0]=0;
        sort(bloom,bloom+cnt);
        ans=1;
        for(int i=1;i<cnt;++i)
            if(bloom[i]!=bloom[i-1]) bloom[ans++]=bloom[i];
        for(int i=ans-1;i>0;--i)
            if(bloom[i]!=bloom[i-1]+1) bloom[ans++]=bloom[i-1]+1;
        sort(bloom,bloom+ans);
        for(int i=ans;i>0;--i)
            if(bloom[i]!=bloom[i-1]+1) bloom[ans++]=bloom[i]-1;
        sort(bloom,bloom+ans);
        //hash处理完毕
        build(0,ans,1);
        for(int i=0;i<n;++i)
            update(binarySearch(num[i].s,ans),binarySearch(num[i].t,ans),0,ans,1);
        for(;m--;)
        {
            scanf("%d",&cnt);
            int pos=binarySearch(cnt,ans);
            printf("%d\n",query(pos,0,ans,1));
        }
    }
    return 0;
}