1、http://acm.hdu.edu.cn/showproblem.php?pid=3308
2、题目大意：
给出一个由n个数字组成的序列，有两种操作，U A B是将A位置的数字替换成B，Q A B是查询A-B区间有多少个递增的数字，也就是求A-B区间的最长上升子序列的长度
这道题目有对区间的查询及对单点的更新，很容易想到线段树，只是在具体实现的时候还是有些困难，具体看代码
要记录区间里的最长的序列长度是多少，用smax来记录。由于最长的连续上升序列可以在区间的左端点，右端点，所以在线段树里增加两个域lmax和rmax，分别表示，在区间的端点处，分别向右向左的最长的满足条件的序列长度。
在更新当前区间的lmax时，如果左子区间的lmax等于左子区间的长度，那么当前区间的lmax就要加上右儿子的lmax。更新当前区间的rmax时同理。
另外在查询的时候，要注意，有可能当前区间的lmax大于mid-st+1，所以要在其中取一个较小值，同理rmax有可能大于ed-mid。
3、题目：
LCIS
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3479 Accepted Submission(s): 1554
Problem Description
Given n integers.
You have two operations:
U A B: replace the Ath number by B. (index counting from 0)
Q A B: output the length of the longest consecutive increasing subsequence (LCIS) in [a, b].
You have two operations:
U A B: replace the Ath number by B. (index counting from 0)
Q A B: output the length of the longest consecutive increasing subsequence (LCIS) in [a, b].
Input
T in the first line, indicating the case number.
Each case starts with two integers n , m(0<n,m<=10 5).
The next line has n integers(0<=val<=10 5).
The next m lines each has an operation:
U A B(0<=A,n , 0<=B=10 5)
OR
Q A B(0<=A<=B< n).
Each case starts with two integers n , m(0<n,m<=10 5).
The next line has n integers(0<=val<=10 5).
The next m lines each has an operation:
U A B(0<=A,n , 0<=B=10 5)
OR
Q A B(0<=A<=B< n).
Output
For each Q, output the answer.
Sample Input
1 10 10 7 7 3 3 5 9 9 8 1 8 Q 6 6 U 3 4 Q 0 1 Q 0 5 Q 4 7 Q 3 5 Q 0 2 Q 4 6 U 6 10 Q 0 9
Sample Output
1 1 4 2 3 1 2 5
Author
shǎ崽
Source
HDOJ Monthly Contest – 2010.02.06
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4、AC代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 400005
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int lmax[N],rmax[N],smax[N];
int a[N];
void pushUp(int d,int m,int rt)
{lmax[rt]=lmax[rt<<1];rmax[rt]=rmax[rt<<1|1];smax[rt]=max(smax[rt<<1],smax[rt<<1|1]);if(a[m]<a[m+1]){if(lmax[rt<<1]==(d-(d>>1)))lmax[rt]+=lmax[rt<<1|1];if(rmax[rt<<1|1]==(d>>1))rmax[rt]+=rmax[rt<<1];smax[rt]=max(smax[rt],rmax[rt<<1]+lmax[rt<<1|1]);}
}
void build(int l,int r,int rt)
{if(l==r){scanf("%d",&a[l]);lmax[rt]=1;rmax[rt]=1;smax[rt]=1;return ;}int m=(l+r)>>1;build(lson);build(rson);pushUp(r-l+1,m,rt);
}
void update(int b,int c,int l,int r,int rt)
{if(l==r){a[b]=c;return ;}int m=(l+r)>>1;if(b<=m)update(b,c,lson);//错在if(r>m)elseupdate(b,c,rson);pushUp(r-l+1,m,rt);
}
int query(int L,int R,int l,int r,int rt)
{if(L<=l && R>=r){return smax[rt];}int m=(l+r)>>1;int ans=-1;if(L<=m)ans=max(ans,query(L,R,lson));if(R>m)ans=max(ans,query(L,R,rson));if(a[m]<a[m+1]){int tmpR=lmax[rt<<1|1]+m;int tmpL=m-rmax[rt<<1]+1;ans=max(ans,min(R,tmpR)-max(L,tmpL)+1);}return ans;
}
int main()
{int t,b,c,n,m;char ch[3];scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);build(0,n-1,1);for(int i=1;i<=m;i++){scanf("%s%d%d",ch,&b,&c);//printf("%c %d %d\n",ch[0],b,c);if(ch[0]=='Q'){int ans=query(b,c,0,n-1,1);printf("%d\n",ans);}else{update(b,c,0,n-1,1);}}}return 0;
}