hdu 4325 Flowers

Problem F

 Time Limit: 4000/2000 MS (Java/Others)     Memory Limit: 65536/65536 K (Java/Others)

	Total Submission(s): 332    Accepted Submission(s): 143	

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 Si and Ti (1 <= Si <= Ti <= 10^9), means i-th flower will be blooming at time [Si, Ti].
In the next M lines, each line contains an integer Ti, 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.

 

SampleInput

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

 

SampleOutput

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

//水水噢
//区间覆盖问题，求某个区间被覆盖几次
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define N 100000
#define lson l,m,k<<1
#define rson m+1,r,k<<1|1
using namespace std;
int st[N<<2];
void build(int l,int r,int k)
{
    st[k]=0;
    if(l==r)
     return;
    int m=(l+r)>>1;
    build(lson);
    build(rson);
}
void update(int &L,int &R,int l,int r,int k)
{
    if(L<=l&&R>=r)
    {
        st[k]+=1;
        return ;
    }
    int m=(l+r)>>1;
    if(L<=m) update(L,R,lson);
    if(R>m)  update(L,R,rson);
}
void query(int &i,int l,int r,int k)
{
    if(l==r)
    {
        printf("%d\n",st[k]);
        return;
    }
    if(st[k])
    {
        st[k<<1]+=st[k];
        st[k<<1|1]+=st[k];
        st[k]=0;
    }
    int m=(l+r)>>1;
    if(i<=m) query(i,lson);
    else query(i,rson);
}
int main()
{
    int T,t=1;
    int i,n,m;
    int s,e;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        build(1,N,1);
        printf("Case #%d:\n",t++);
        for(i=1;i<=n;i++)
         {
             scanf("%d%d",&s,&e);
             update(s,e,1,N,1);
         }
         while(m--)
         {
             scanf("%d",&s);
             query(s,1,N,1);
         }
    }
    return 0;
}