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For a sequence S1,S2,...,SN, and a pair of integers (i, j), if 1 <= i <= j <= N and Si < Si+1 < Si+2 <...< Sj-1 < Sj, then the sequence Si,Si+1,...,Sj is a CIS (Continuous Increasing Subsequence). The longest CIS of a sequence is called the LCIS (Longest Continuous Increasing Subsequence).
In this problem, we will give you a sequence first, and then some “add” operations and some “query” operations. An add operation adds a value to each member in a specified interval. For a query operation, you should output the length of the LCIS of a specified interval.
The first line of the input is an integer T, which stands for the number of test cases you need to solve.
Every test case begins with two integers N, Q, where N is the size of the sequence, and Q is the number of queries. S1,S2,...,SN are specified on the next line, and then Q queries follow. Every query begins with a character ‘a’ or ‘q’. ‘a’ is followed by three integers L, R, V, meaning that add V to members in the interval [L, R] (including L, R), and ‘q’ is followed by two integers L, R, meaning that you should output the length of the LCIS of interval [L, R].
For every test case, you should output "Case #k:" on a single line first, where k indicates the case number and starts at 1. Then for every ‘q’ query, output the answer on a single line. See sample for more details.
1
5 6
0 1 2 3 4
q 1 4
a 1 2 -10
a 1 1 -6
a 5 5 -4
q 2 3
q 4 4
Case #1:
4
2
1
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 100010; 4 struct node { 5 int ret,lv,rv,lsum,rsum,lazy; 6 } tree[maxn<<2]; 7 void pushup(int v,int k) { 8 tree[v].lsum = tree[v<<1].lsum; 9 tree[v].rsum = tree[v<<1|1].rsum; 10 tree[v].lv = tree[v<<1].lv; 11 tree[v].rv = tree[v<<1|1].rv; 12 if(tree[v].lsum == k - (k>>1) && tree[v<<1].rv < tree[v<<1|1].lv) 13 tree[v].lsum += tree[v<<1|1].lsum; 14 if(tree[v].rsum == (k>>1) && tree[v<<1].rv < tree[v<<1|1].lv) 15 tree[v].rsum += tree[v<<1].rsum; 16 tree[v].ret = max(tree[v<<1].ret,tree[v<<1|1].ret); 17 if(tree[v<<1].rv < tree[v<<1|1].lv) 18 tree[v].ret = max(tree[v].ret,tree[v<<1].rsum + tree[v<<1|1].lsum); 19 } 20 void pushdown(int v,int k) { 21 if(tree[v].lazy) { 22 tree[v<<1].lazy += tree[v].lazy; 23 tree[v<<1|1].lazy += tree[v].lazy; 24 tree[v<<1].lv += tree[v].lazy; 25 tree[v<<1].rv += tree[v].lazy; 26 tree[v<<1|1].lv += tree[v].lazy; 27 tree[v<<1|1].rv += tree[v].lazy; 28 tree[v].lazy = 0; 29 } 30 } 31 void build(int L,int R,int v) { 32 tree[v].lazy = 0; 33 if(L == R) { 34 tree[v].lsum = tree[v].rsum = 1; 35 scanf("%d",&tree[v].rv); 36 tree[v].lv = tree[v].rv; 37 tree[v].ret = 1; 38 return; 39 } 40 int mid = (L + R)>>1; 41 build(L,mid,v<<1); 42 build(mid+1,R,v<<1|1); 43 pushup(v,R - L + 1); 44 } 45 void update(int L,int R,int lt,int rt,int val,int v) { 46 if(lt <= L && rt >= R) { 47 tree[v].lazy += val; 48 tree[v].lv += val; 49 tree[v].rv += val; 50 return; 51 } 52 pushdown(v,R - L + 1); 53 int mid = (L + R)>>1; 54 if(lt <= mid) update(L,mid,lt,rt,val,v<<1); 55 if(rt > mid) update(mid+1,R,lt,rt,val,v<<1|1); 56 pushup(v,R - L + 1); 57 } 58 int query(int L,int R,int lt,int rt,int v) { 59 if(lt <= L && rt >= R) return tree[v].ret; 60 pushdown(v,R - L + 1); 61 int ret = 0,mid = (L + R)>>1; 62 if(lt <= mid) ret = max(ret,query(L,mid,lt,rt,v<<1)); 63 if(rt > mid) ret = max(ret,query(mid+1,R,lt,rt,v<<1|1)); 64 if(lt <= mid && rt > mid && tree[v<<1].rv < tree[v<<1|1].lv) 65 ret = max(ret,min(mid - lt + 1,tree[v<<1].rsum) + min(rt - mid,tree[v<<1|1].lsum)); 66 pushup(v,R - L + 1); 67 return ret; 68 } 69 int main() { 70 int T,n,m,x,y,val,cs = 1; 71 char op[3]; 72 scanf("%d",&T); 73 while(T--) { 74 scanf("%d %d",&n,&m); 75 printf("Case #%d:\n",cs++); 76 build(1,n,1); 77 while(m--) { 78 scanf("%s%d%d",op,&x,&y); 79 if(op[0] == ‘a‘) { 80 scanf("%d",&val); 81 update(1,n,x,y,val,1); 82 } else if(op[0] == ‘q‘) 83 printf("%d\n",query(1,n,x,y,1)); 84 } 85 } 86 return 0; 87 } 88 /* 89 1 90 5 6 91 0 1 2 3 4 92 q 1 4 93 a 1 2 -10 94 a 1 1 -6 95 a 5 5 -4 96 q 2 3 97 q 4 4 98 99 Case #1: 100 4 101 2 102 1 103 */
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原文地址：http://www.cnblogs.com/crackpotisback/p/4466542.html
