Super Mario

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

题意：给你一个序列，以及m个询问，每次查询[l,r]里面小于等于h的数的个数

分析：离线保存序列和询问，排序后将序列从小到大插入，然后对于询问，用树状数组或者线段树查询区间和就行了。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;

const int MAXN = 1E5+100;
struct Query
{
    int l,r;
    int h;
    int id;
}Q[MAXN];
pair<int,int>p[MAXN];
int F[MAXN];
int n,m;
void update(int x,int val)
{
    while(x<=n)
    {
        F[x]+=val;
        x+=x&-x;
    }
}
int query(int x)
{
    int res=0;
    while(x>0)
    {
        res+=F[x];
        x-=x&-x;
    }
    return res;
}
bool cmp(const Query& aa,const Query& bb)
{
    return aa.h<bb.h;
}
int ans[MAXN];

int main()
{
    int T;
    int iCase=0;
    scanf("%d",&T);
    while(T--)
    {
        iCase++;
        scanf("%d%d",&n,&m);
        memset(F,0,sizeof(F));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&p[i].first);
            p[i].second=i;
        }
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&Q[i].l,&Q[i].r,&Q[i].h);
            Q[i].id=i,Q[i].l++,Q[i].r++;
        }
        sort(Q+1,Q+m+1,cmp);
        sort(p+1,p+n+1);
        int i=1,j=1;
        while(j<=m)
        {
            while(i<=n)
            {
                if(p[i].first>Q[j].h) break;
                update(p[i].second,1);
                i++;
            }
            while(j<=m)
            {
                if(i<=n&&Q[j].h>=p[i].first) break;
                ans[Q[j].id]=query(Q[j].r)-query(Q[j].l-1);
                j++;
            }
        }
        printf("Case %d:\n",iCase);
        for(int i=1;i<=m;i++) printf("%d\n",ans[i]);
    }
    return 0;
}