nyoj----522 Interval （简单树状数组）

Interval

时间限制： 2000  ms  |  内存限制： 65535  KB

难度： 4

 

描述

There are n(1 <= n <= 100000) intervals [ai, bi] and m(1 <= m <= 100000) queries, -100000 <= ai <= bi <= 100000 are integers.

Each query contains an integer xi(-100000 <= x <= 100000). For each query, you should answer how many intervals convers xi.

 

输入

The first line of input is the number of test case.
For each test case,
two integers n m on the first line,  
then n lines, each line contains two integers ai, bi;
then m lines, each line contains an integer xi.

输出

m lines, each line an integer, the number of intervals that covers xi.

样例输入

2

3 4

1 3

1 2

2 3

0

1

2

3

1 3

0 0

-1

0

1

样例输出

0

2

3

2

0

1

0

上传者

ACM_赵铭浩

 

代码：

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<stdlib.h>

 4 #define maxn 200005   //整体平移100001个单位

 5 #define lowbit(x) ((x)&(-x))

 6 int aa[maxn+5];

 7 void ope(int x,int val)

 8 {

 9      x+=100001;

10     while(x<=maxn)

11     {

12         aa[x]+=val;

13         x+=lowbit(x);

14     }

15 }

16 long long getsum(int x)

17 {

18    long long 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 test,nn,m,i,a,b;

29     scanf("%d",&test);

30     while(test--)

31     {

32       scanf("%d%d",&nn,&m);

33       memset(aa,0,sizeof(aa));

34       for(i=0;i<nn;i++)

35       {

36        scanf("%d%d",&a,&b);

37        ope(a,1);

38        ope(b+1,-1);

39       }

40       for(i=0;i<m;i++)

41       {

42        scanf("%d",&a);

43        a+=100001;

44        printf("%I64d\n",getsum(a));

45       }

46     }

47   return 0;

48 }

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