HDOJ 4325 —— 树状数组 + 离散化

Flowers

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

Total Submission(s): 1981    Accepted Submission(s): 974


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

 

Recommend

zhoujiaqi2010

 
离线查询，所以先把端点和查询点离散化。然后根据树状数组的插线问点即可。

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>

using namespace std;
//#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
//#define mid ((l + r) >> 1)
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 1000000 + 50;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
#define mod 1000000007
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pi;
///#pragma comment(linker, "/STACK:102400000,102400000")
int c[MAXN] , fl[MAXN << 2] , q[MAXN];
struct Node
{
int st , en;
}site[MAXN];
int lowbit(int x)
{
return x & ( -x);
}
void update(int x , int value , int n)
{
while(x <= n)
{
c[x] += value;
x += lowbit(x);
}
}
int getsum(int x)
{
int sum = 0;
while(x > 0)
{
sum += c[x];
x -= lowbit(x);
}
return sum;
}
int main()
{
//ios::sync_with_stdio(false);
#ifdef Online_Judge
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif // Online_Judge
int t , n , nn, m;
scanf("%d" , &t);
FORR(kase , 1 , t)
{

scanf("%d%d" , &n , &m);
int num = 0;
clr(site , 0);
clr(fl , 0);
clr(c , 0);
FOR(i , 0 , n)
{
scanf("%d%d" ,&site[i].st , &site[i].en);
fl[num++] = site[i].st;
fl[num++] = site[i].en;
}
FOR(i , 0 , m)
{
scanf("%d" , &q[i]);
fl[num++] = q[i];
}
//        outstars
sort(fl , fl + num);
num = unique(fl  , fl + num) - fl;
FOR(i ,0 , n)
{
int x = lower_bound(fl , fl + num , site[i].st) - fl;
int y = lower_bound(fl , fl + num , site[i].en) - fl;
update(x + 1 , 1 , num),update(y + 2, -1 , num);
}

printf("Case #%d:\n", kase);
FOR(i ,0 , m)
{
int x = lower_bound(fl , fl + num , q[i]) - fl + 1;
printf("%d\n" ,getsum(x));
}

}
return 0;
}