HDUOJ-----2852 KiKi's K-Number（樹狀陣列+二分）
KiKi's K-Number
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2393 Accepted Submission(s): 1101
Problem Description
For the k-th number, we all should be very familiar with it. Of course,to kiki it is also simple. Now Kiki meets a very similar problem, kiki wants to design a container, the container is to support the three operations. Push: Push a given element e to container Pop: Pop element of a given e from container Query: Given two elements a and k, query the kth larger number which greater than a in container; Although Kiki is very intelligent, she can not think of how to do it, can you help her to solve this problem?
Input
Input some groups of test data ,each test data the first number is an integer m (1 <= m <100000), means that the number of operation to do. The next m lines, each line will be an integer p at the beginning, p which has three values: If p is 0, then there will be an integer e (0 <e <100000), means press element e into Container. If p is 1, then there will be an integer e (0 <e <100000), indicated that delete the element e from the container If p is 2, then there will be two integers a and k (0 <a <100000, 0 <k <10000),means the inquiries, the element is greater than a, and the k-th larger number.
Output
For each deletion, if you want to delete the element which does not exist, the output "No Elment!". For each query, output the suitable answers in line .if the number does not exist, the output "Not Find!".
Sample Input
5 0 5 1 2 0 6 2 3 2 2 8 1 7 0 2 0 2 0 4 2 1 1 2 1 2 2 1 3 2 1 4
Sample Output
No Elment! 6 Not Find! 2 2 4 Not Find!
Source
2009 Multi-University Training Contest 4 - Host by HDU
這道題是一題，題意是要找a的第k大的數，由於想到樹狀陣列，也有這麼一種解法，所以就拿來練練手，首先採用傳統的樹狀陣列上升模式，進行
累加和，不過這裡的和表示的是比a比大的個數，所以在我們查詢的時候只要將要查詢的的數sum(val)-sum(a)==k來進行二分，當然由於資料是離散的，而我們這裡有無法進行離散化處理，因此我們只需要加入一個輔助陣列lab[]，進行判斷就可以了.....這道題，第一次遇到可算是想了一段時間，畢竟是個菜鳥....做出來還是很幸福的說....
程式碼如下：
1 #include<stdio.h>
2 #include<string.h>
3 #include<stdlib.h>
4 #define maxn 100000
5 #define lowbit(x) ((x)&(-x))
6 int aa[maxn+3];
7 int lab[maxn+3];
8 void ope(int x,int val)
9 {
10 while(x<=maxn)
11 {
12 aa[x]+=val;
13 x+=lowbit(x);
14 }
15 }
16 int cont(int x)
17 {
18 int ans=0;
19 while(x>0)
20 {
21 ans+=aa[x];
22 x-=lowbit(x);
23 }
24 return ans;
25 }
26 int main()
27 {
28 int n,p,a,k;
29 int left,right,mid,res;
30 bool tag=true;
31 // freopen("test.in","r",stdin);
32 //freopen("test.out","w",stdout);
33 while(scanf("%d",&n)!=EOF)
34 {
35 memset(aa,0,sizeof(aa));
36 memset(lab,0,sizeof(lab));
37 while(n--)
38 {
39 scanf("%d",&p);
40 if(p==2)
41 {
42 scanf("%d%d",&a,&k);
43 left=a, right=maxn;
44 tag=true;
45 while(left<=right)
46 {
47 mid=left+(right-left)/2;
48 res=cont(mid)-cont(a);
49 if(res<k)
50 left=mid+1;
51 else if(res-lab[mid]>=k)
52 right=mid-1;
53 else
54 if(res>=k&&res-lab[mid]<k)
55 {
56 printf("%dn",mid);
57 tag=false;
58 break;
59 }
60 }
61 if(tag)
62 printf("Not Find!n");
63
64 }
65 else
66 {
67 scanf("%d",&a);
68 if(p==1)
69 {
70 if(lab[a]==0)
71 printf("No Elment!n");
72 else
73 {
74 lab[a]--;
75 ope(a,-1); //1代表insert
76 }
77 }
78 else
79 {
80 lab[a]++;
81 ope(a,1); //0代表elete
82 }
83 }
84 }
85 }
86 return 0;
87 }並給出一些資料來幫助驗證吧....
1 5
2 0 5
3 1 2
4 0 6
5 2 3 2
6 2 8 1
7 7
8 0 2
9 0 2
10 0 4
11 2 1 1
12 2 1 2
13 2 1 3
14 2 1 4
15 12
16 0 2
17 0 2
18 0 4
19 0 5
20 1 2
21 0 6
22 2 3 2
23 2 8 1
24 2 1 1
25 2 1 2
26 2 1 3
27 2 1 4