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 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.

 

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;

}