Super Mario

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9501    Accepted Submission(s): 4017

 

Problem 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

Source

2012 ACM/ICPC Asia Regional Hangzhou Online

題目連線：HDU4417 Super Mario

題解：求靜態區間[l,r]中小於等於k的數的數量，主席樹模板題。注意，題目中區間從0開始編號

AC的C++程式碼：

#include<iostream>
#include<algorithm>

using namespace std;
const int N=100010;
int a[N],b[N],rt[N*20],ls[N*20],rs[N*20],sum[N*20];

int id;
void build(int &o,int l,int r)
{
	o=++id;
	sum[o]=0;
	if(l==r) return;
	int m=(l+r)>>1;
	build(ls[o],l,m);
	build(rs[o],m+1,r);
}

void update(int &o,int l,int r,int last,int p)
{
	o=++id;
	ls[o]=ls[last];
	rs[o]=rs[last];
	sum[o]=sum[last]+1;
	if(l==r) return;
	int m=(l+r)>>1;
	if(p<=m)
	  update(ls[o],l,m,ls[last],p);
	else
	  update(rs[o],m+1,r,rs[last],p);
}
//查詢區間[L,R]內小於等於k的個數 
int query(int s,int e,int l,int r,int k)
{
	if(b[r]<=k)
	  return sum[e]-sum[s];
	if(l==r)
	  return 0;
	int m=(l+r)>>1;
	int sum=0;
	if(b[m+1]<=k)
	  sum+=query(rs[s],rs[e],m+1,r,k);
	sum+=query(ls[s],ls[e],l,m,k);
	return sum;
}

int main()
{
	int T,n,q,l,r,k;
	scanf("%d",&T);
	for(int t=1;t<=T;t++){
		scanf("%d%d",&n,&q);
		for(int i=1;i<=n;i++){
			scanf("%d",&a[i]);
			b[i]=a[i];
		}
		sort(b+1,b+1+n);
		int size=unique(b+1,b+1+n)-(b+1);
		id=0;
		build(rt[0],1,size);
		for(int i=1;i<=n;i++){
			a[i]=lower_bound(b+1,b+1+size,a[i])-b;
			update(rt[i],1,size,rt[i-1],a[i]);
		}
		printf("Case %d:\n",t);
		while(q--){
			scanf("%d%d%d",&l,&r,&k);
			l++,r++; //題中是從0開始編號的，因此使它加一從1開始編號 
			int num=query(rt[l-1],rt[r],1,size,k);
			printf("%d\n",num);
		}
	}
	return 0;
}