HDU 4417-Super Mario-线段树+离线

Description

Mario is world-famous plumber. His “burly” figure and amazing jumping ability reminded in our memory. Now the poor princess is in trouble again and Mario needs to save his lover. We regard the road to the boss’s castle as a line (the length is n), on every integer point i there is a brick on height hi. Now the question is how many bricks in [L, R] Mario can hit if the maximal height he can jump is H.

Input

The first line follows an integer T, the number of test data.
For each test data:
The first line contains two integers n, m (1 <= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the number of queries.
Next line contains n integers, the height of each brick, the range is [0, 1000000000].
Next m lines, each line contains three integers L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.)

Output

For each case, output "Case X: " (X is the case number starting from 1) followed by m lines, each line contains an integer. The ith integer is the number of bricks Mario can hit for the ith query.

Sample Input

1
10 10
0 5 2 7 5 4 3 8 7 7 
2 8 6
3 5 0
1 3 1
1 9 4
0 1 0
3 5 5
5 5 1
4 6 3
1 5 7
5 7 3

Sample Output

Case 1:
4
0
0
3
1
2
0
1
5
1

题目大意：

线段树的区间查询，查询区间[l,r]中<=h的数的个数。

核心思想：

线段树+离线。

代码如下：

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=1e5+20;
struct node{
	int x,id;
}a[N];
struct Node{
	int l,r,h,id,chu;
}que[N];
struct tnode{
	int l,r,sum;
}tr[N<<2];
bool cmp0(node p,node q)
{
	return p.x<q.x;
}
bool cmp1(Node p,Node q)
{
	return p.h<q.h;
}
bool cmp2(Node p,Node q)
{
	return p.id<q.id;
}
void pushup(int m)
{
	tr[m].sum=tr[m<<1].sum+tr[m<<1|1].sum;
	return;
}
void build(int m,int l,int r)
{
	tr[m].l=l;
	tr[m].r=r;
	if(l==r)
	{
		tr[m].sum=0;
		return;
	}
	int mid=(l+r)>>1;
	build(m<<1,l,mid);
	build(m<<1|1,mid+1,r);
	pushup(m);
	return;
}
void update(int m,int x)
{
	if(tr[m].l==x&&tr[m].r==x)
	{
		tr[m].sum=1;
		return;
	}
	int mid=(tr[m].l+tr[m].r)>>1;
	if(x<=mid)
		update(m<<1,x);
	else
		update(m<<1|1,x);
	pushup(m);
	return;
}
int query(int m,int l,int r)
{
	if(tr[m].l==l&&tr[m].r==r)
		return tr[m].sum;
	int mid=(tr[m].l+tr[m].r)>>1;
	if(r<=mid)
		return query(m<<1,l,r);
	if(l>mid)
		return query(m<<1|1,l,r);
	return query(m<<1,l,mid)+query(m<<1|1,mid+1,r);
}
int main()
{
	int T,ca=0;
	cin>>T;
	while(T--)
	{
		int n,m;
		//输入并建树 
		scanf("%d%d",&n,&m);
		build(1,1,n);
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&a[i].x);
			a[i].id=i;
		}
		//排序，小的先进树 
		sort(a+1,a+1+n,cmp0);
		//离线处理 
		for(int i=1;i<=m;i++)
		{
			scanf("%d%d%d",&que[i].l,&que[i].r,&que[i].h);
			que[i].l++;
			que[i].r++;
			que[i].id=i;
		}
		sort(que+1,que+1+m,cmp1);
		//边更新边离线查询 
		int k=1;
		for(int i=1;i<=m;i++)
		{
			while(k<=n&&a[k].x<=que[i].h)
			{
				update(1,a[k].id);
				k++;
			}
			que[i].chu=query(1,que[i].l,que[i].r);
		}
		//把顺序sort回来，输出 
		sort(que+1,que+1+m,cmp2);
		printf("Case %d:
",++ca);
		for(int i=1;i<=m;i++)
			printf("%d
",que[i].chu);
	}
	return 0;
}