（线段树）UESTC 360-Another LCIS

For a sequence S1,S2,⋯,SNS1,S2,⋯,SN, and a pair of integers (i,j)(i,j), if 1≤i≤j≤N1≤i≤j≤N and Si<Si+1<Si+2<⋯<Sj−1<SjSi<Si+1<Si+2<⋯<Sj−1<Sj, then the sequence Si,Si+1,⋯,SjSi,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.

Input

The first line of the input is an integer TT, which stands for the number of test cases you need to solve.

Every test case begins with two integers NN, QQ, where NN is the size of the sequence, and QQ is the number of queries. S1,S2,⋯,SNS1,S2,⋯,SN are specified on the next line, and then QQ queries follow. Every query begins with a character a or q. a is followed by three integers LL, RR, VV, meaning that add VV to members in the interval [L,R][L,R] (including LL, RR), and q is followed by two integers LL, RR, meaning that you should output the length of the LCIS of interval [L,R][L,R].

T≤10T≤10;

1≤N,Q≤1000001≤N,Q≤100000;

1≤L≤R≤N1≤L≤R≤N;

−10000≤S1,S2,⋯,SN,V≤10000−10000≤S1,S2,⋯,SN,V≤10000.

Output

For every test case, you should output Case #k: on a single line first, where kkindicates the case number and starts at 11. Then for every q query, output the answer on a single line. See sample for more details.

Sample Input

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

Sample Output

Case #1:
4
2
1

线段树维护各个左闭右开区间。需要注意的是区间合并的条件，为方便判断，每个结点记录这个区间左右两侧的值，方便进行比较能否合并。每次查询、修改的时候，没完全在目标区间内的时候
不要忘记将之前未传下去的改变传递下去。
  1 #include <iostream>
  2 //#include<bits/stdc++.h>
  3 #include <stack>
  4 #include <queue>
  5 #include <map>
  6 #include <set>
  7 #include <cstdio>
  8 #include <cstring>
  9 #include <algorithm>
 10 #include <math.h>
 11 using namespace std;
 12 typedef long long ll;
 13 typedef unsigned long long ull;
 14 const int MAX=1e5+5;
 15 struct node
 16 {
 17     int left,right,mid;//分别表示从区间左侧开始最长的单调子序列，区间中最长，以右侧结尾的最长单调子序列长度
 18     int x,y;//区间左侧的值，右侧的值
 19     int num;
 20 }st[10*MAX];
 21 //int col[10*MAX];
 22 void pushup(int l,int r,int k)
 23 {
 24     st[k].left=st[2*k].left;
 25     st[k].right=st[2*k+1].right;
 26     st[k].mid=max(st[2*k].mid,st[2*k+1].mid);
 27     st[k].x=st[2*k].x;
 28     st[k].y=st[2*k+1].y;
 29     if(st[2*k].y<st[2*k+1].x)
 30     {
 31         st[k].mid=max(st[k].mid,st[2*k].right+st[2*k+1].left);
 32         if(st[2*k].left==(l+r)/2-l)
 33             st[k].left=st[2*k].left+st[2*k+1].left;
 34         if(st[2*k+1].right==r-(l+r)/2)
 35             st[k].right=st[2*k+1].right+st[2*k].right;
 36     }
 37 }
 38 void pushdown(int k)
 39 {
 40     if(st[k].num)
 41     {
 42         st[2*k].num+=st[k].num;
 43         st[2*k+1].num+=st[k].num;
 44         st[2*k].x+=st[k].num;
 45         st[2*k].y+=st[k].num;
 46         st[2*k+1].x+=st[k].num;
 47         st[2*k+1].y+=st[k].num;
 48         st[k].num=0;
 49     }
 50 }
 51 void init(int l,int r,int k)
 52 {
 53     if(l+1==r)
 54     {
 55         st[k].left=st[k].right=st[k].mid=1;
 56         scanf("%d",&st[k].num);
 57         st[k].x=st[k].y=st[k].num;
 58         return;
 59     }
 60     st[k].num=0;
 61     init(l,(l+r)/2,2*k);
 62     init((l+r)/2,r,2*k+1);
 63     pushup(l,r,k);
 64 }
 65 void update(int ul,int ur,int l,int r,int k,int z)//[ul,ur)是目标区间，[l,r)是当下区间,z是加的值
 66 {
 67     if(l>=ul&&r<=ur)
 68     {
 69         st[k].num+=z;
 70         st[k].x+=z;
 71         st[k].y+=z;
 72         return;
 73     }
 74     pushdown(k);
 75     if(l+1==r)
 76         return;
 77     int m=(l+r)/2;
 78     if(ul<m)
 79         update(ul,ur,l,m,2*k,z);
 80     if(ur>m)
 81         update(ul,ur,m,r,2*k+1,z);
 82     pushup(l,r,k);
 83 }
 84 int query(int ql,int qr,int l,int r,int k)//[ql,qr)是查询区间，[l,r)是当下区间
 85 {
 86     if(l>=ql&&r<=qr)
 87         return st[k].mid;
 88     pushdown(k);
 89     int m=(l+r)/2;
 90     int maxn;
 91     if(ql<m)
 92     {
 93         if(qr<=m)
 94             return query(ql,qr,l,m,2*k);
 95         else
 96             {
 97                 maxn=max(query(ql,qr,l,m,2*k),query(ql,qr,m,r,2*k+1));
 98                 if(st[2*k].y<st[2*k+1].x)
 99                 {
100                     maxn=max(maxn,min(m-ql,st[2*k].right)+min(qr-m,st[2*k+1].left));
101                 }
102                 return maxn;
103             }
104     }
105     else
106         return query(ql,qr,m,r,2*k+1);
107 }
108 int t;
109 int N,Q;
110 int main()
111 {
112     scanf("%d",&t);
113     char tem[10];
114     int l,r,zhi;
115     for(int i=1;i<=t;i++)
116     {
117         printf("Case #%d:\n",i);
118         scanf("%d %d",&N,&Q);
119         init(1,N+1,1);
120         for(int j=1;j<=Q;j++)
121         {
122             scanf("%s",tem);
123             if(tem[0]=='q')
124             {
125                 scanf("%d %d",&l,&r);
126                 printf("%d\n",query(l,r+1,1,N+1,1));
127             }
128             else
129             {
130                 scanf("%d %d %d",&l,&r,&zhi);
131                 update(l,r+1,1,N+1,1,zhi);
132             }
133         }
134     }
135 }


 


 






            

            
posted @ 
2017-02-22 14:41 
perplex 
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