hdu4325 樹狀陣列+離散化

life4711發表於2015-02-06
陣列

http://acm.hdu.edu.cn/showproblem.php?pid=4325

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 Si and Ti (1 <= Si <= Ti <= 10^9), means i-th flower will be blooming at time [Si, Ti].
In the next M lines, each line contains an integer Ti, 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






/**
hdu 4325 樹狀陣列+離散化
題目大意：給定n朵花，每朵花開放的時間段，然後m個詢問：ti時間點有多少花正在盛開
解題思路：樹狀陣列裸體，不過要離散化花開放的時間
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=300005;

int n,m,k;
int C[maxn],a[maxn];

struct note
{
    int x,id;
    bool operator <(const note &other)const
    {
        return x<other.x;
    }
}p[maxn];

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

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

void update(int pos,int value)
{
    while(pos<=k)
    {
        C[pos]+=value;
        pos+=lowbit(pos);
    }
}
int main()
{
    int T,tt=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        int nn=n*2;
        int mm=nn+m;
        for(int i=0;i<mm;i++)
        {
            scanf("%d",&p[i].x);
            p[i].id=i;
        }
        sort(p,p+mm);
        k=1;
        a[p[0].id]=k;
        for(int i=1;i<mm;i++)
        {
            if(p[i].x==p[i-1].x)
            {
                a[p[i].id]=k;
            }
            else
                a[p[i].id]=++k;
        }
        memset(C,0,sizeof(C));
        for(int i=0;i<nn-1;i+=2)
        {
            update(a[i],1);
            update(a[i+1]+1,-1);
        }
        printf("Case #%d:\n",++tt);
        for(int i=nn;i<mm;i++)
        {
            printf("%d\n",sum(a[i]));
        }
    }
    return 0;
}









相關文章