Flowers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3150    Accepted Submission(s): 1549

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.  

 

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  

 

Author BJTU  

 

Source 2012 Multi-University Training Contest 3  

/*
给你每一朵话的开花时间段，询问你某一时刻的开花数量
*/

/*
重新定义树状数组的意义，不再是前i个数的和，而是第i个位置的数值
*/
/*
明显数据会爆的，我去.....数据太水了
*/
#include<iostream>
#include<string.h>
#include<stdio.h>
#define N 100010
using namespace std;
int c
,T
;
int t;
int n,m;
int lowbit(int x)
{
return x&(-x);
}
void update(int x,int val)
{
while(x<=N)
{
c[x]+=val;
x+=lowbit(x);
}
}
int getsum(int x)
{
int s=0;
while(x>0)
{
s+=c[x];
x-=lowbit(x);
}
return s;
}
int main()
{
//freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
scanf("%d",&t);
for(int Case=1;Case<=t;Case++)
{
memset(c,0,sizeof c);
scanf("%d%d",&n,&m);
//cout<<"n="<<n<<" "<<"m="<<m<<endl;
int si,ti;
for(int i=0;i<n;i++)
{
scanf("%d%d",&si,&ti);
//cout<<si<<" "<<ti<<endl;

update(si,1);

// for(int j=si;j<=ti;j++)
// update(j,1);
update(ti+1,-1);
}
for(int i=0;i<m;i++)
scanf("%d",&T[i]);
printf("Case #%d:\n",Case);
// for(int i=1;i<10;i++)
// printf("%d\n",getsum(i));
for(int i=0;i<m;i++)
{
//cout<<"T[i]="<<T[i]<<" "<<"T[i]-1="<<T[i]-1<<endl;
printf("%d\n",getsum(T[i]));
}
}
}