Poj Buy Tickets (line segment tree)

Buy Tickets

Time Limit:4000 MS		Memory Limit:65536 K
Total Submissions:7802		Accepted:3710

Description

Railway tickets were difficult to buy around the Lunar New Year in China, so we must get up early and join a long queue...

The Lunar New Year was approaching, but unluckily the Little Cat still had schedules going here and there. now, he had to travel by train to Mianyang, Sichuan Province for the winter camp selection of the national team of Olympus in ICS ICs.

It was one o 'clock a.m. and dark outside. chill wind from the northwest did not scare off the people in the queue. the cold night gave the Little Cat a shiver. why not find a problem to think about? That was none the less better than freezing to death!

People kept jumping the queue. since it was too dark around, such moves wocould not be discovered even by the people adjacent to the queue-jumpers. "If every person in the queue is assigned an integral value and all the information about those who have jumped
The queue and where they stand after queue-jumping is given, can I find out the final order of people in the queue ?" Thought the Little Cat.

Input

There will be several test cases in the input. Each test case consistsN+ 1 lines whereN(1 ≤N≤ 200,000) is given in the first line of the test case. The nextNLines contain the pairs of valuesPosiAndValiIn
The increasing orderI(1 ≤I≤N). For eachI, The ranges and meaningsPosiAndValiAre as follows:

• Posiε [0,I−1]-I-Th person came to the queue and stood right behindPosi-Th person in the queue. The booking office was considered the 0th person and the person at the front of the queue
  Was considered the first person in the queue.
• Valiε [0, 32767]-I-Th person was assigned the valueVali.

There no blank lines between test cases. Proceed to the end of input.

Output

For each test cases, output a single line of space-separated integers which are the values of people in the order they stand in the queue.

Sample Input

40 771 511 332 6940 205231 192431 38900 31492

Sample Output

77 33 69 5131492 20523 3890 19243

Hint

The figure below shows how the Little Cat found out the final order of people in the queue described in the first test case of the sample input.

Source

POJ Monthly -- 2006.05.28, Zhu, Zeyuan this question has been thinking for a long time, and finally can not help but read the ideas of others and then found that it is a problem of processing input, people think of http://acm.zju.edu.cn/onlinejudge/showProblem.do? ProblemId = 3563 the problem of deformation in this query set is also handled backwards, which solves the defect that the query set has not been deleted. This may be a inspiration for future questions, but now there is not much feeling back to this question: because the last person to be inserted must be in a fixed position (because no one will squeeze him back later) so after this person's position is arranged, "discard" the position that has been arranged before determining the next person's position. In this way, insert it to the I position, that is, it is not difficult to understand the principle of searching for the I + 1 Vacancy. The key is how to design the search function, this requires you to think carefully and independently. I combined the search and modification during processing, because every time I search for the leaves, another problem occurred while returning to the root/*. The result of the stack timed out because you need to input the data first and then process it backwards, it will be okay if you change back to the array. Remember not to use stl in the future. The efficiency is lower */The Code is as follows:

# Include <stdio. h> int p [200005]; int v [200005]; int ans [200005]; struct Node {int l, r, num;} tree [1000000]; void Build (int n, int x, int y) {tree [n]. l = x; tree [n]. r = y; tree [n]. num = (y-x + 1); // stores the int mid = (x + y)/2; if (x = y) {return;} Build (2 * n, x, mid); Build (2 * n + 1, mid + 1, y);} void Modify (int n, int p, int v) {int l = tree [n]. l; int r = tree [n]. r; int mid = (l + r)/2; if (l = r) {tree [n]. Num = 0; ans [l] = v; return;} if (tree [2 * n]. num> = p) Modify (2 * n, p, v); else Modify (2 * n + 1, p-tree [2 * n]. num, v); tree [n]. num = tree [2 * n]. num + tree [2 * n + 1]. num;} int main () {int n; int I; while (scanf ("% d", & n )! = EOF) {Build (1, 1, n); for (I = 0; I <n; I ++) {scanf ("% d ", & p [I], & v [I]) ;}for (I = n-1; I> = 0; I --) {Modify (1, p [I] + 1, v [I]) ;}for (I = 1; I <n; I ++) {printf ("% d ", ans [I]);} printf ("% d \ n", ans [n]);} return 0 ;}