HDU4325--Flowers--树状数组,离散化

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

很简单的一道树状数组的改段求点的模版题,基本思想是:tree[]为树状数组,N个元素,tree[i]存放的是i至N被加了多少,

所以,修改[l,r]区间值的操作就变为了,

update(l,1);

update(r+1,-1);

查询某点的操作就变为了查询此点的前缀和

需要注意的是,此题目的数据范围太大,需要进行离散化
1 (关于离散化,会再写一篇博客)
```c++
#include
#include
#include
#include
#include
using namespace std;
const int M=100100;
int tree[M];
int q[M];
int N=0;
map ls;
struct node
{
int le;
int ri;
}a[M];
int lowbits(int x)
{
return x&(-x);
}
void update(int i,int val)
{
for(;i0;i-=lowbits(i)){
sum+=tree[i];
}
return sum;
}
void init()
{
memset(tree,0,sizeof(tree));
ls.clear();
}
int main()
{
freopen("data.in","r",stdin);
int m,n;
int t;
int l,r;
int tem;
int casn=1;
cin>>t;
while(t--){
cout>n>>m;
for(int i=0;i>l>>r;
if(ls.find(l)==ls.end()) ls.insert(make_pair(l,1));
if(ls.find(r)==ls.end()) ls.insert(make_pair(r,1));
a[i].le=l;a[i].ri=r;
}
for(int i=0;i>l;
if(ls.find(l)==ls.end()) ls.insert(make_pair(l,1));
q[i]=l;
}
int i=1;
for(map::iterator ite=ls.begin();ite!=ls.end();ite++){
ite->second=i++;
}
N=i-1;
for(int i=0;i