HDU 4325 Flowers 树状数组+离散化

Flowers

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

分析

　　我觉得你只要不傻，应该都能看出来这是个树状数组区间修改单点查询的板子题，但数据好像有点大，开数组是肯定开不下的，而最多就1e5朵花，所以在1e9之间有很多数字都被浪费掉了，所以使用离散化，离散化时记得将相同的数标记成一样的数字，还有询问也要离散化。

　　然鹅这道题我很好奇的试了试，没有离散化直接让开1e5的数组然后跑树状数组，让我非常吃惊的是，它A了！ HDU数据好水 出题人的意思肯定是让你用离散化，所以……看代码吧

　　

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e5+5;
int tr[N+5],a[N+5];
struct Node{
    int id,x;
    bool operator <(const Node &A)const {
        return x<A.x;//方便处理相等的情况
    }
}p[N+5];
//树状数组开始
int lowbit(int x){
    return x&(-x);
}
void Add(int x,int w){
    while(x<=N){
        tr[x]+=w;
        x+=lowbit(x);
    }
}
int Sum(int x){
    int ans=0;
    while(x){
        ans+=tr[x];
        x-=lowbit(x);
    }
    return ans;
}
//树状数组结束
int main(){
    int n,cas=0;
    scanf("%d",&n);
    while(n--){
        printf("Case #%d:
",++cas);
        int totn,totm;
        scanf("%d%d",&totn,&totm);
        totn*=2;
        totm+=totn;
        for(int i=1;i<=totm;i++){
            scanf("%d",&p[i].x);
            p[i].id=i;
            //所有数据一次读完，并记录i的值，因为结构体排序后顺序被打乱
        }
        sort(p+1,p+totm+1);
        a[p[1].id]=1;
        int k=1;
        for(int i=2;i<=totm;i++){
            if(p[i].x==p[i-1].x)a[p[i].id]=k;//相等的值肯定也要离散成一样的呀
            else a[p[i].id]=++k;
        }
        memset(tr,0,sizeof(tr));
        for(int i=1;i<=totn;i+=2){
            Add(a[i],1);
            Add(a[i+1]+1,-1);
        }
        for(int i=totn+1;i<=totm;i++){
            printf("%d
",Sum(a[i]));
        }
    }
    return 0;
}