hdu4417 樹狀陣列（求指定區間比指定數小的數的個數）

life4711發表於2015-04-05
陣列

http://acm.hdu.edu.cn/showproblem.php?pid=4417

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




/**
hdu4417  樹狀陣列（求指定區間比指定數小的數的個數）
http://blog.csdn.net/wmn_wmn/article/details/8034181
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=300005;

int C[maxn],ans[maxn];
int n,m;

int lowbit(int x)
{
    return x&(-x);
}

int sum(int x)
{
    int ret=0;
    while(x>0)
    {
        ret+=C[x];
        x-=lowbit(x);
    }
    return ret;
}

void add(int x,int d)
{
    while(x<=n)
    {
        C[x]+=d;
        x+=lowbit(x);
    }
}

struct node
{
    int id,num;
    bool operator < (const node &other) const
    {
        return num < other.num;
    }
}a[maxn];
struct note
{
    int l,r,id,value;
    bool operator < (const note &other)const
    {
        return value <other.value;
    }
}b[maxn];

int main()
{
    int T,tt=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
        {
            int x;
            scanf("%d",&a[i].num);
            a[i].id=i+1;
        }
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d",&b[i].l,&b[i].r,&b[i].value);
            b[i].l++;
            b[i].r++;
            b[i].id=i+1;
        }
        sort(a,a+n);
        sort(b,b+m);
        memset(C,0,sizeof(C));
        int k=0;
        for(int i=0;i<m;i++)
        {
            while(k<n&&a[k].num<=b[i].value)
            {
                add(a[k].id,1);
                k++;
            }
            ans[b[i].id]=sum(b[i].r)-sum(b[i].l-1);
        }
        printf("Case %d:\n",++tt);
        for(int i=1;i<=m;i++)
        {
            printf("%d\n",ans[i]);
        }
    }
    return 0;
}









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