HDU1556(树状数组)

Color the ball

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17044 Accepted Submission(s): 8513

Problem Description

N个气球排成一排，从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的“小飞鸽"牌电动车从气球a开始到气球b依次给每个气球涂一次颜色。但是N次以后lele已经忘记了第I个气球已经涂过几次颜色了，你能帮他算出每个气球被涂过几次颜色吗？

Input

每个测试实例第一行为一个整数N,(N <= 100000).接下来的N行，每行包括2个整数a b(1 <= a <= b <= N)。
当N = 0，输入结束。

Output

每个测试实例输出一行，包括N个整数，第I个数代表第I个气球总共被涂色的次数。

Sample Input

1 1

2 2

3 3

1 1

1 2

1 3

Sample Output

1 1 1

3 2 1

树状数组入门，区间更新，单点查询

 1 //2016.8.10
 2 //树状数组，区间更新，单点查询
 3 #include<iostream>
 4 #include<cstdio>
 5 #include<cstring>
 6 
 7 using namespace std;
 8 
 9 int arr[100005], n;//arr[i]表示i管辖的范围内数字之和
10 
11 int lowbit(int x){return x&(-x);}
12 
13 int add(int pos, int tt)
14 {
15     for(int i = pos; i <= n; i+=lowbit(i))
16       arr[i] += tt;
17     return 0;
18 }
19 
20 int query(int pos)
21 {
22     int sum = 0; 
23     for(int i = pos; i > 0; i-=lowbit(i))
24       sum+=arr[i];
25     return sum;
26 }
27 
28 int main()
29 {
30     int a, b;
31     while(cin>>n)
32     {
33         memset(arr, 0, sizeof(arr));
34         for(int i = 0; i < n; i++)
35         {
36             scanf("%d%d", &a, &b);
37             add(a, 1);
38             add(b+1, -1);
39         }
40         for(int i = 1; i <= n; i++)
41           if(i == n)
42             cout<<query(i)<<endl;
43           else cout<<query(i)<<" ";
44     }
45 
46     return 0;
47 }

 1 // 2018-03-28
 2 
 3 import java.util.*;
 4     
 5 public class Main {
 6 
 7     public static void main(String[] args) {
 8         Scanner cin = new Scanner(System.in);
 9         
10         int n;
11         while(cin.hasNext()) {
12             n = cin.nextInt();
13             if(n == 0)break;
14             BinaryIndexedTrees bit = new BinaryIndexedTrees(n);
15             int l, r;
16             for(int i = 0; i < n; i++) {
17                 l = cin.nextInt();
18                 r = cin.nextInt();
19                 bit.updata(l, r, 1);
20             }
21             
22             for(int i = 1; i <= n; i++) {
23                 if(i == n)System.out.printf("%d 
", bit.query(i));
24                 else System.out.printf("%d ", bit.query(i));
25             }
26         }
27     }
28 }
29     
30 class BinaryIndexedTrees {
31     /*
32      * 区间更新，单点查询
33      */
34     int n;
35     int [] e;
36     
37     BinaryIndexedTrees(int _n) {
38         this.n = _n;
39         e = new int[n+1];
40     }
41     
42     // x+lowbit(x)表示到达x的父节点
43     // x-lowbit(x)表示到达x点管辖区间的下个区间的管辖点
44     int lowbit(int x) {
45         return x&(-x);
46     }
47     
48     void add(int pos, int val) {
49         for(int i = pos; i <= n; i+=lowbit(i))
50             e[i] += val;
51     }
52     
53     int query(int pos) {
54         int sum = 0;
55         for(int i = pos; i > 0; i-=lowbit(i))
56             sum += e[i];
57         return sum;
58     }
59     
60     // 给[l, r]区间内的每个数加上val
61     void updata(int l, int r, int val) {
62         add(l, val);
63         add(r+1, -val);
64         
65     }
66 }