Codeforces 295A Greg and Array

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A. Greg and Array

time limit per test 　　1.5 seconds

memory limit per test 　　256 megabytes

input 　　standard input

output　　standard output

Greg has an array a = a₁, a₂, ..., a_n and m operations. Each operation looks as: l_i, r_i, d_i, (1 ≤ l_i ≤ r_i ≤ n). To apply operation i to the array means to increase all array elements with numbers l_i, l_i + 1, ..., r_i by value d_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 numbers x_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 contains n integers: a₁, a₂, ..., a_n (0 ≤ a_i ≤ 10⁵) — the initial array.

Next m lines contain operations, the operation number i 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 number i 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 in C++. It is preferred to use the cin, cout streams of the %I64d specifier.

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
 
分析
线段树。
离线预处理所有Query，统计各operation的次数。
区间Insert，注意使用lazy-tag，点Query答案。
写法
要维护两棵线段树，可并做一棵。
 
这是我第一次写的，TLE on test 24

 #include<bits/stdc++.h>
 using namespace std;
 const int MAX_N=1e5+;
 typedef long long ll;
 
 struct op{
     int l, r;
     ll v;
 }o[MAX_N];
 
 ll cnt[MAX_N], a[MAX_N];
 
 struct Node{
     int l, r;
     ll v;
     int mid(){return (l+r)>>;}
 }T[MAX_N<<];
 
 void Build(int id, int l, int r){
     T[id].l=l, T[id].r=r, T[id].v=;
     if(l==r) return;
     int mid=T[id].mid();
     Build(id<<, l, mid);
     Build(id<<|, mid+, r);
 }
 void Insert(int id, int l, int r, ll v){
     Node &now=T[id];
     if(now.l>=l&&now.r<=r){
         if(~now.v) now.v+=v;
         else{
             Insert(id<<, l, r, v);
             Insert(id<<|, l, r, v);
         }
     }
     else{
         Node &lch=T[id<<], &rch=T[id<<|];
         if(~now.v) lch.v=rch.v=now.v, now.v=-; //ERROR-PRONE
         int mid=now.mid();
         if(l<=mid) Insert(id<<, l, r, v);
         if(r>mid) Insert(id<<|, l, r, v);
         if(lch.v==rch.v) now.v=lch.v;
     }
 }
 
 void Qurery(int id, ll *a){
     Node &now=T[id];
     if(~now.v)
         for(int i=now.l; i<=now.r; i++) a[i]+=now.v;
     else{
         Qurery(id<<, a);
         Qurery(id<<|, a);
     }
 }
 
 int main(){
     //freopen("in", "r", stdin);
     int N, M, K;
     scanf("%d%d%d", &N, &M, &K);
     for(int i=; i<=N; i++) scanf("%lld", a+i);
     for(int i=; i<=M; i++)
         scanf("%d%d%lld", &o[i].l, &o[i].r, &o[i].v);
     Build(, , M);
     int l, r;
     while(K--){
         scanf("%d%d", &l, &r);
         Insert(, l, r, );
     }
     Qurery(, cnt);
     Build(, , N);
     for(int i=; i<=M; i++)
         if(cnt[i])
             Insert(, o[i].l, o[i].r, o[i].v*cnt[i]);
     Qurery(, a);
     for(int i=; i<=N; i++)
         printf("%lld ", a[i]);
     puts("");
     return ;
 }

上面的代码没有lazy-tag或者说我设置的lazy-tag没起到相应的作用。我的考虑是设置一个tag，最后求答案时可不必细分到每个叶子节点，但是这种优化对降低Insert的复杂度没有太大帮助，而Insert是最耗时的，因而总的复杂度还是没降下来。
AC的姿势

 #include<bits/stdc++.h>
 using namespace std;
 const int MAX_N=1e5+;
 typedef long long ll;
 
 struct op{
     int l, r, v;
 }o[MAX_N];
 
 ll cnt[MAX_N], a[MAX_N];
 
 struct Node{
     int l, r;
     ll v;
     int mid(){return (l+r)>>;}
 }T[MAX_N<<];
 
 void Build(int id, int l, int r){
     T[id].l=l, T[id].r=r, T[id].v=;
     if(l==r) return;
     int mid=T[id].mid();
     Build(id<<, l, mid);
     Build(id<<|, mid+, r);
 }
 void Insert(int id, int l, int r, ll v){
     Node &now=T[id];
     if(now.l>=l&&now.r<=r) now.v+=v;
     else{
         Node &lch=T[id<<], &rch=T[id<<|];
         if(now.v)
             lch.v+=now.v, rch.v+=now.v, now.v=;
         int mid=now.mid();
         if(l<=mid) Insert(id<<, l, r, v);
         if(r>mid) Insert(id<<|, l, r, v);
     }
 }
 
 void Qurery(int id, ll *a){
     Node &now=T[id];
     if(now.l==now.r) a[now.l]+=now.v;
     else{
         Node &lch=T[id<<], &rch=T[id<<|];
         if(now.v)
             lch.v+=now.v, rch.v+=now.v;
         Qurery(id<<, a);
         Qurery(id<<|, a);
     }
 }
 
 int main(){
     //freopen("in", "r", stdin);
     int N, M, K;
     scanf("%d%d%d", &N, &M, &K);
     for(int i=; i<=N; i++) scanf("%lld", a+i);
     for(int i=; i<=M; i++)
         scanf("%d%d%lld", &o[i].l, &o[i].r, &o[i].v);
     Build(, , M);
     int l, r;
     while(K--){
         scanf("%d%d", &l, &r);
         Insert(, l, r, );
     }
     Qurery(, cnt);
     Build(, , N);
     for(int i=; i<=M; i++)
         if(cnt[i]&&o[i].v)
             Insert(, o[i].l, o[i].r, o[i].v*cnt[i]);
     Qurery(, a);
     for(int i=; i<=N; i++)
         printf("%lld ", a[i]);
     puts("");
     return ;
 }