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+;
 int tr[N+],a[N+];
 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+];
 //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=;
     while(x){
         ans+=tr[x];
         x-=lowbit(x);
     }
     return ans;
 }
 //树状数组结束
 int main(){
     int n,cas=;
     scanf("%d",&n);
     while(n--){
         printf("Case #%d:\n",++cas);
         int totn,totm;
         scanf("%d%d",&totn,&totm);
         totn*=;
         totm+=totn;
         for(int i=;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+,p+totm+);
         a[p[].id]=;
         int k=;
         for(int i=;i<=totm;i++){
             if(p[i].x==p[i-].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,,sizeof(tr));
         for(int i=;i<=totn;i+=){
             Add(a[i],);
             Add(a[i+]+,-);
         }
         for(int i=totn+;i<=totm;i++){
             printf("%d\n",Sum(a[i]));
         }
     }
     return ;
 }