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 paper. 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.
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.
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 %I64d specifier.
3 3 3
1 2 3
1 2 1
1 3 2
2 3 4
1 2
1 3
2 3
9 18 17
1 1 1
1
1 1 1
1 1
2
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
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
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int MAX_N=1e5+10; 4 typedef long long ll; 5 6 struct op{ 7 int l, r; 8 ll v; 9 }o[MAX_N]; 10 11 ll cnt[MAX_N], a[MAX_N]; 12 13 struct Node{ 14 int l, r; 15 ll v; 16 int mid(){return (l+r)>>1;} 17 }T[MAX_N<<2]; 18 19 void Build(int id, int l, int r){ 20 T[id].l=l, T[id].r=r, T[id].v=0; 21 if(l==r) return; 22 int mid=T[id].mid(); 23 Build(id<<1, l, mid); 24 Build(id<<1|1, mid+1, r); 25 } 26 void Insert(int id, int l, int r, ll v){ 27 Node &now=T[id]; 28 if(now.l>=l&&now.r<=r){ 29 if(~now.v) now.v+=v; 30 else{ 31 Insert(id<<1, l, r, v); 32 Insert(id<<1|1, l, r, v); 33 } 34 } 35 else{ 36 Node &lch=T[id<<1], &rch=T[id<<1|1]; 37 if(~now.v) lch.v=rch.v=now.v, now.v=-1; //ERROR-PRONE 38 int mid=now.mid(); 39 if(l<=mid) Insert(id<<1, l, r, v); 40 if(r>mid) Insert(id<<1|1, l, r, v); 41 if(lch.v==rch.v) now.v=lch.v; 42 } 43 } 44 45 void Qurery(int id, ll *a){ 46 Node &now=T[id]; 47 if(~now.v) 48 for(int i=now.l; i<=now.r; i++) a[i]+=now.v; 49 else{ 50 Qurery(id<<1, a); 51 Qurery(id<<1|1, a); 52 } 53 } 54 55 int main(){ 56 //freopen("in", "r", stdin); 57 int N, M, K; 58 scanf("%d%d%d", &N, &M, &K); 59 for(int i=1; i<=N; i++) scanf("%lld", a+i); 60 for(int i=1; i<=M; i++) 61 scanf("%d%d%lld", &o[i].l, &o[i].r, &o[i].v); 62 Build(1, 1, M); 63 int l, r; 64 while(K--){ 65 scanf("%d%d", &l, &r); 66 Insert(1, l, r, 1); 67 } 68 Qurery(1, cnt); 69 Build(1, 1, N); 70 for(int i=1; i<=M; i++) 71 if(cnt[i]) 72 Insert(1, o[i].l, o[i].r, o[i].v*cnt[i]); 73 Qurery(1, a); 74 for(int i=1; i<=N; i++) 75 printf("%lld ", a[i]); 76 puts(""); 77 return 0; 78 }
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
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int MAX_N=1e5+10; 4 typedef long long ll; 5 6 struct op{ 7 int l, r, v; 8 }o[MAX_N]; 9 10 ll cnt[MAX_N], a[MAX_N]; 11 12 struct Node{ 13 int l, r; 14 ll v; 15 int mid(){return (l+r)>>1;} 16 }T[MAX_N<<2]; 17 18 void Build(int id, int l, int r){ 19 T[id].l=l, T[id].r=r, T[id].v=0; 20 if(l==r) return; 21 int mid=T[id].mid(); 22 Build(id<<1, l, mid); 23 Build(id<<1|1, mid+1, r); 24 } 25 void Insert(int id, int l, int r, ll v){ 26 Node &now=T[id]; 27 if(now.l>=l&&now.r<=r) now.v+=v; 28 else{ 29 Node &lch=T[id<<1], &rch=T[id<<1|1]; 30 if(now.v) 31 lch.v+=now.v, rch.v+=now.v, now.v=0; 32 int mid=now.mid(); 33 if(l<=mid) Insert(id<<1, l, r, v); 34 if(r>mid) Insert(id<<1|1, l, r, v); 35 } 36 } 37 38 void Qurery(int id, ll *a){ 39 Node &now=T[id]; 40 if(now.l==now.r) a[now.l]+=now.v; 41 else{ 42 Node &lch=T[id<<1], &rch=T[id<<1|1]; 43 if(now.v) 44 lch.v+=now.v, rch.v+=now.v; 45 Qurery(id<<1, a); 46 Qurery(id<<1|1, a); 47 } 48 } 49 50 int main(){ 51 //freopen("in", "r", stdin); 52 int N, M, K; 53 scanf("%d%d%d", &N, &M, &K); 54 for(int i=1; i<=N; i++) scanf("%lld", a+i); 55 for(int i=1; i<=M; i++) 56 scanf("%d%d%lld", &o[i].l, &o[i].r, &o[i].v); 57 Build(1, 1, M); 58 int l, r; 59 while(K--){ 60 scanf("%d%d", &l, &r); 61 Insert(1, l, r, 1); 62 } 63 Qurery(1, cnt); 64 Build(1, 1, N); 65 for(int i=1; i<=M; i++) 66 if(cnt[i]&&o[i].v) 67 Insert(1, o[i].l, o[i].r, o[i].v*cnt[i]); 68 Qurery(1, a); 69 for(int i=1; i<=N; i++) 70 printf("%lld ", a[i]); 71 puts(""); 72 return 0; 73 }