Interval (Nanyang oj522) (tree-like array)

Interval time limit: -Ms | Memory Limit:65535KB Difficulty:4

Describe

There is n (1 <= n <= 100000) intervals [AI, bi] and m (1 <= m <= 100000) queries, -100000 <= ai <= bi &l t;= 100000 is integers. Each query contains an integer XI ( -100000 <= x <= 100000). For each query, you should answer how many intervals Convers XI.

Input

The first line of input was the number of the

test case.
For each test case,
Integers n m on the first line,
Then n lines, each line contains the integers AI, bi;
Then M. Lines, each line contains an integer XI.

Output

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

Sample input

23 41 31 22 301231 30 0-101

Sample output

0232010

 
/* a tree-like array application, The number of occurrences of the statistic in the given interval. Tle cried, max size: 200002 is ac,200005 is tle!!! Note the negative number is converted to a positive range, or else it will be tle!!! */#include <stdio.h> #include <string.h> #define MAX 200002int a[max];int lowbit (int N) {return n& (-n);} void asd (int i,int M) {while (I<=max) {a[i]+=m;i+=lowbit (i);}} int sum (int j) {int sum=0;while (j>0) {sum+=a[j];j-=lowbit (j);} return sum;} int main () {int i,n,m,t,t,k,p;scanf ("%d", &t), while (t--) {memset (a,0,sizeof (a)); scanf ("%d%d", &m,&n); while (m--) {scanf ("%d%d", &k,&t), ASD (k+100001,1), ASD (t+100002,-1);} while (n--) {scanf ("%d", &p);p rintf ("%d\n", Sum (p+100001));}} return 0;} 
 
 
 

Interval (Nanyang oj522) (tree-like array)