hdoj--4325--Flowers(线段树+二分)

Flowers

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

Total Submission(s): 2585    Accepted Submission(s): 1271

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.

 

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

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define MAX 100000
struct tree
{
	int l,r;
	int m;
}num[MAX*10];
void build(int node,int l,int r)
{
	num[node].l=l;
	num[node].r=r;
	num[node].m=0;
	if(l==r)
	return ;
	int mid=(l+r)/2;
	build(node*2,l,mid);
	build(node*2+1,mid+1,r);
}
void updata(int node,int l,int r)
{
	if(num[node].l==l&&num[node].r==r)
	{
		num[node].m++;
		return ;
	}
	int mid=(num[node].l+num[node].r)/2;
	if(r<=mid)
	updata(node*2,l,r);
	else if(l>mid)
	updata(node*2+1,l,r);
	else
	{
		updata(node*2,l,mid);
		updata(node*2+1,mid+1,r);
	}
}
int query(int node,int l,int r)
{
	if(num[node].l==l&&num[node].r==r)
	{
		return num[node].m;
	}
	int mid=(num[node].l+num[node].r)/2;
	if(r<=mid)
	return query(node*2,l,r)+num[node].m;
	else
	{
		if(l>mid)
		return query(node*2+1,l,r)+num[node].m;
		else
		return query(node*2,l,mid)+query(node*2+1,mid+1,r);
	}
}
int main()
{
	int t;
	int Case=1;
	scanf("%d",&t);
	while(t--)
	{
		int m,n,x,y,z;
		scanf("%d%d",&n,&m);
		build(1,1,MAX);
		while(n--)
		{
			scanf("%d%d",&x,&y);
			updata(1,x,y);
		}
		printf("Case #%d:\n",Case++);
		while(m--)
		{
			scanf("%d",&z);
			printf("%d\n",query(1,z,z));
		}
	}
	return 0;
}

醉了醉了，二分都可以

<pre name="code" class="cpp">#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define MAX  100010
int a[MAX],b[MAX];
int main()
{
	int t;
	int Case=1;
	scanf("%d",&t);
	while(t--)
	{
		int m,n,x,y,z;
		scanf("%d%d",&n,&m);
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		for(int i=0;i<n;i++)
		scanf("%d%d",&a[i],&b[i]);
		sort(a,a+n);sort(b,b+n);
		printf("Case #%d:\n",Case++);
		for(int i=0;i<m;i++)
		{
			scanf("%d",&z);
			x=upper_bound(a,a+n,z)-a;
			y=lower_bound(b,b+n,z)-b;
			printf("%d\n",x-y);
		}
	}
	return 0;
}