POJ2528 Mayor & amp; #39; s posters line segment tree

POJ2528 Mayor & #39; s posters line segment tree
Mayor's posters

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:49723		Accepted:14419

Description

The citizens of Bytetown, AB, cocould not stand that the candidates in the mayoral election campaign have been placing their electoral posters at all places at their whim. the city councel has finally decided to build an electoral wall for placing the posters and introduce the following rules:
Every candidate can place exactly one poster on the wall.
All posters are of the same height equal to the height of the wall; the width of a poster can be any integer number of bytes (byte is the unit of length in Bytetown ).
The wall is divided into segments and the width of each segment is one byte.
Each poster must completely cover a contiguous number of wall segments.
They have built a wall 10000000 bytes long (such that there is enough place for all candidates ). when the electoral campaign was restarted, the candidates were placing their posters on the wall and their posters differed widely in width. moreover, the candidates started placing their posters on wall segments already occupied by other posters. everyone in Bytetown was curous whose posters will be visible (entirely or in part) on the last day before elections.
Your task is to find the number of visible posters when all the posters are placed given the information about posters 'size, their place and order of placement on the electoral wall.

Input

The first line of input contains a number c giving the number of cases that follow. the first line of data for a single case contains number 1 <=n <= 10000. the subsequent n lines describe the posters in the order in which they were placed. the I-th line among the n lines contains two integer numbers li and ri which are the number of the wall segment occupied by the left end and the right end of the I-th poster, respectively. we know that for each 1 <= I <= n, 1 <= li <= ri <= 10000000. after the I-th poster is placed, it entirely covers all wall segments numbered li, li + 1 ,..., ri.

Output

For each input data set print the number of visible posters after all the posters are placed.

The picture below has strates the case of the sample input.

Sample Input

151 42 68 103 47 10

Sample Output

4

First, consider discretization. The correct discretization method is as follows:
That is, each line segment is represented by the two ends of the unit interval covered by this line segment. Medium: Line Segment ① is represented as []; line segment ② is represented as []; the line segment ③ is [3, 4].
After discretization, We can insert all the line segments and query them again. That is, the dyeing problem maintains such information cover [k]: whether the interval corresponding to k is completely colored by the same color. The colors are 1, 2, 3 ,..., n indicates that a bool array vis is used in the first query of all intervals with 0 to query the entire interval. If there is any color, set vis of this color to true, finally, the number of true in vis is the answer.

# Include
 
  
# Include
  
   
# Include using namespace std; // In fact, after discretization, the total unit interval can reach 20000 + 19999 at most. // Therefore, cover to maxn * 2*2*4 const int maxn = 10000 + 10; int cover [maxn * 16], x [maxn], y [maxn], s [maxn * 2]; unsigned short Hash [10000005]; // unsigned short is enough, if short is not enough, revoid pushDown (int k) {if (cover [k]> = 0) {cover [k * 2] = cover [k * 2 + 1] = cover [k]; cover [k] =-1 ;}} void Insert (int, int B, int c, int k, int l, int r) {if (a <= l & r <= B) {cover [k] = c; return ;} pushDown (k); int m = (l + r)/2; if (a <= m) Insert (a, B, c, k * 2, l, m ); if (B> m) Insert (a, B, c, k * 2 + 1, m + 1, r);} bool vis [maxn]; void ask (int k, int l, int r) {if (cover [k]> = 0) {vis [cover [k] = true; return;} if (l = r) return; pushDown (k); int m = (l + r)/2; ask (k * 2, l, m); ask (k * 2 + 1, m + 1, r) ;}int main () {int I, T, n; scanf (% d, & T); while (T --) {scanf (% d, & n ); memset (Hash, 0, sizeof (Hash); int tot = 0; for (I = 0; I