CF 179(div2) C(线段树 || 扫描法 )

C. Greg and Array

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Greg has an array a = a₁, a₂, ..., a_n andm operations. Each operation looks as: l_i, r_i, d_i, (1 ≤ l_i ≤ r_i ≤ n). To apply operationi to the array means to increase all array elements with numbersl_i, l_i + 1, ..., r_i by valued_i.

Greg wrote down k queries on a piece of paper. Each query has the following form:x_i,y_i,(1 ≤ x_i ≤ y_i ≤ m). That means that one should apply operations with numbersx_i, x_i + 1, ..., y_i 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 containsn integers: a₁, a₂, ..., a_n(0 ≤ a_i ≤ 10⁵) — the initial array.

Next m lines contain operations, the operation numberi is written as three integers: l_i, r_i, d_i, (1 ≤ l_i ≤ r_i ≤ n),(0 ≤ d_i ≤ 10⁵).

Next k lines contain the queries, the query numberi is written as two integers: x_i, y_i, (1 ≤ x_i ≤ y_i ≤ m).

The numbers in the lines are separated by single spaces.

Output

On a single line print n integers a₁, a₂, ..., a_n — 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 inC++. It is preferred to use the cin, cout streams of the %I64d specifier.

Sample test(s)

Input

3 3 31 2 31 2 11 3 22 3 41 21 32 3

Output

9 18 17

Input

1 1 111 1 11 1

Output

2

Input

4 3 61 2 3 41 2 12 3 23 4 41 21 32 31 21 32 3

Output

 

这道题有两种做法，第一种也是最常想到的维护两颗线段树。其实这道题有另一种O（n）的算法，因为我们最后只查询一次，所以是不需要线段树的。关键是找到类似线段树lazy的另一种数据结构。用一个count数组记录，a是累加后的结果，关键代码就是这句了

for(int i = 1;i <= m;i++){    countn[l[i]-1] += d[i];countn[r[i]] -= d[i];}sum = 0;for(int i = 0;i <= n;i++){     a[i] += sum;sum += countn[i];}

读者自己动手模拟一遍就能看出里面的思路了

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define LL long longusing namespace std;const int maxn = 100100;LL a[maxn],l[maxn],r[maxn],d[maxn];LL countn[maxn],ll[maxn],rr[maxn];int main(){    int n,m,k;    while(scanf("%d%d%d",&n,&m,&k) != EOF){        for(int i = 1;i <= n;i++)            cin >> a[i];        for(int i = 1;i <= m;i++)            cin >> l[i] >> r[i] >> d[i];        memset(countn,0,sizeof(countn));        for(int i = 0;i < k;i++){            int teml,temr;            cin >> teml >> temr;            countn[teml-1] += 1;countn[temr] -= 1;        }        LL sum = 0;        for(int i = 0;i <= m;i++){            d[i] *= sum;            sum += countn[i];        }        memset(countn,0,sizeof(countn));        for(int i = 1;i <= m;i++){            countn[l[i]-1] += d[i];countn[r[i]] -= d[i];        }        sum = 0;        for(int i = 0;i <= n;i++){            a[i] += sum;sum += countn[i];        }        for(int i = 1;i < n;i++) cout << a[i] << ' ';        cout << a[n] << endl;    }    return 0;}