Codeforces 295A Greg and Array

传送门

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 ;
 }