题目链接：https://csacademy.com/contest/round-45/task/remove-update/
题目大意：N个数初始值为0，现在给出Q个操作(l，r，x)，每个操作将区间[l，r]的值加x。倘若除去其中一次更新，那么整个数列的最大值就会改变。问除去一次更新能达到的数列最大值的最小值是多少。
解题思路：线段树区间更新，维护最大值。先进行Q此操作，之后枚举每个操作，执行操作(l，r，-x)，记录最大值，之后执行(l，r，x)还原。
代码：
1 const int maxn = 1e6 + 5; 2 struct node{ 3 int v, inc; 4 }; 5 node tree[maxn]; 6 int n, q; 7 int al[maxn], ar[maxn], ax[maxn]; 8 9 void build(int l, int r, int k){ 10 tree[k].v = tree[k].inc = 0; 11 if(l == r) return; 12 int mid = (l + r) >> 1, lc = k << 1, rc = k << 1 | 1; 13 build(l, mid, lc); 14 build(mid + 1, r, rc); 15 } 16 void update(int ul, int ur, int x, int l, int r, int k){ 17 if(ul <= l && ur >= r){ 18 tree[k].inc += x; 19 return; 20 } 21 if(ul > r || ur < l) return; 22 int mid = (l + r) >> 1, lc = k << 1, rc = k << 1 | 1; 23 tree[k].v = tree[k].v + tree[k].inc; 24 tree[lc].inc += tree[k].inc; tree[rc].inc += tree[k].inc; 25 tree[k].inc = 0; 26 update(ul, ur, x, l, mid, lc); 27 update(ul, ur, x, mid + 1, r, rc); 28 tree[k].v = max(tree[lc].v + tree[lc].inc, tree[rc].v + tree[rc].inc); 29 } 30 31 void solve(){ 32 build(1, n, 1); 33 for(int i = 1; i <= q; i++){ 34 update(al[i], ar[i], ax[i], 1, n, 1); 35 } 36 int ans = inf; 37 for(int i = 1; i <= q; i++){ 38 update(al[i], ar[i], -ax[i], 1, n, 1); 39 ans = min(ans, tree[1].v + tree[1].inc); 40 update(al[i], ar[i], ax[i], 1, n, 1); 41 } 42 printf("%d ", ans); 43 } 44 int main(){ 45 scanf("%d %d", &n, &q); 46 for(int i = 1; i <= q; i++){ 47 int l, r, x; 48 scanf("%d %d %d", &l, &r, &x); 49 al[i] = l; ar[i] = r; ax[i] = x; 50 } 51 solve(); 52 }
题目：
Remove Update
Time limit: 1000 ms
Memory limit: 256 MB
Memory limit: 256 MB
You have an array AA of size NN. Initially all the elements of AA are equal to 00.
You have QQ updates of the form:
- Given ll, rr and xx (1 leq l leq r leq N1≤l≤r≤N, xx is positive), add xx to A_l, A_{l+1},..., A_rAl,Al+1,...,Ar
You can choose to skip exactly one of the updates. In the end you want the maximum element of AA to be as small as possible.
Standard input
The first line contains two integers NN and QQ.
Each of the next QQ lines contains three integers ll, rr and xx.
Standard output
Print on the first line the smallest maximum value you can get by skipping one update.
Constraints and notes
- 1 leq N leq 10^51≤N≤105
- 1 leq Q leq 10^51≤Q≤105
- 1 leq l leq r leq N1≤l≤r≤N
- 1 leq x leq 10^91≤x≤109
- It is guaranteed that if you perform all the updates A_i leq 10^9Ai≤109
| Input | Output |
|---|---|
4 2 1 3 1 2 4 1 |
1 |
4 3 2 4 2 1 2 2 3 4 1 |
2 |
6 4 2 3 1 4 6 2 3 4 3 1 4 2 |
4 |