Codeforces Round #179 (Div. 2) C Greg and Array

C. Greg and Array
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, ..., ri by value di.

Greg wrote down k queries on a piece of 易做图. Each query has the following form: xi, yi, (1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbers xi, xi + 1, ..., yi to the array.

Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.

Input
The first line contains integers n, m, k (1 ≤ n, m, k ≤ 105). The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 105) — the initial array.

Next m lines contain operations, the operation number i is written as three integers: li, ri, di, (1 ≤ li ≤ ri ≤ n), (0 ≤ di ≤ 105).

Next k lines contain the queries, the query number i is written as two integers: xi, yi, (1 ≤ xi ≤ yi ≤ m).

The numbers in the lines are separated by single spaces.

Output
On a single line print n integers a1, a2, ..., an — the array after executing all the queries. Separate the printed numbers by spaces.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64dspecifier.

Sample test(s)
input
3 3 3
1 2 3
1 2 1
1 3 2
2 3 4
1 2
1 3
2 3
output
9 18 17
input
1 1 1
1
1 1 1
1 1
output
2
input
4 3 6
1 2 3 4
1 2 1
2 3 2
3 4 4
1 2
1 3
2 3
1 2
1 3
2 3
output
5 18 31 20
思路：这个题数据量很大。我用的方法是构造两个线段树来处理首先要知道线段树的一个比较基本的模型，就是线段树成段的更新（一段+C或-C）只要知道这个基本模型，题目可以分解为两不，第一步，利用x,y来构造线段树，判断每个操作出现了多少次然后再把数组a[n]构造为一颗线段树，并且成段更新的时候，由于第一颗线段树已经可以知道当前这个操作执行了多少次，所以可以直接把di*c作为一个增加的量，来成段更新，这个di就是原来的增量，ci是通过第一颗线段树查找（i,i）这个区间出现的次数最后把每一个ai都查找一次，就可以得到答案了。

补充：综合编程 , 其他综合 ,