2014
03-16

# LCIS

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].

T in the first line, indicating the case number.
Each case starts with two integers n , m(0<n,m<=105).
The next line has n integers(0<=val<=105).
The next m lines each has an operation:
U A B(0<=A,n , 0<=B=105)
OR
Q A B(0<=A<=B< n).

T in the first line, indicating the case number.
Each case starts with two integers n , m(0<n,m<=105).
The next line has n integers(0<=val<=105).
The next m lines each has an operation:
U A B(0<=A,n , 0<=B=105)
OR
Q A B(0<=A<=B< n).

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

1
1
4
2
3
1
2
5

~~

#include <iostream>
#include <cstring>
#include <cstdio>
#define lson pos<<1
#define rson pos<<1|1
using namespace std;
const int MAXN=100005;
struct node
{
int l,r;
int msum;
int lsum,rsum;
int mid()
{
return (l+r)>>1;
}
};
node tree[MAXN*4];
int num[MAXN];
inline void pushup(int pos)
{
tree[pos].msum=max(tree[lson].msum,tree[rson].msum);
tree[pos].lsum=tree[lson].lsum;
tree[pos].rsum=tree[rson].rsum;
if(num[tree[lson].r]<num[tree[rson].l])
{
tree[pos].msum=max(tree[pos].msum,tree[lson].rsum+tree[rson].lsum);
if(tree[lson].lsum==tree[lson].r-tree[lson].l+1)
tree[pos].lsum+=tree[rson].lsum;
if(tree[rson].rsum==tree[rson].r-tree[rson].l+1)
tree[pos].rsum+=tree[lson].rsum;
}
}
void build(int l,int r,int pos)
{
tree[pos].l=l;
tree[pos].r=r;
if(l==r)
{
tree[pos].lsum=tree[pos].rsum=tree[pos].msum=1;
return ;
}
int mid=tree[pos].mid();
build(l,mid,lson);
build(mid+1,r,rson);
pushup(pos);
}
void update(int a,int b,int pos)
{
if(tree[pos].l==tree[pos].r)
{
num[a]=b;
return ;
}
int mid=tree[pos].mid();
if(a<=mid)
update(a,b,lson);
else
update(a,b,rson);
pushup(pos);
}
int queryL(int l,int r,int pos)
{
if(r-l+1>=tree[pos].lsum)
return tree[pos].lsum;
else
return r-l+1;
}
int queryR(int l,int r,int pos)
{
if(r-l+1>=tree[pos].rsum)
return tree[pos].rsum;
else
return r-l+1;
}
int query(int l,int r,int pos)
{
if(l==tree[pos].l&&r==tree[pos].r)
return tree[pos].msum;
int mid=tree[pos].mid();
if(r<=mid)
return query(l,r,lson);
else if(l>mid)
return query(l,r,rson);
else
{
int res=0;
if(num[tree[lson].r]<num[tree[rson].l])
res=max(res,queryR(l,mid,lson)+queryL(mid+1,r,rson));
res=max(res,query(l,mid,lson));
res=max(res,query(mid+1,r,rson));
return res;
}
}

int main()
{
int n,m,t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
int i;
for(i=1;i<=n;i++)
scanf("%d",&num[i]);
build(1,n,1);
char word[2];
for(i=1;i<=m;i++)
{
int a,b;
scanf("%s%d%d",word,&a,&b);
if(word[0]=='U')
update(a+1,b,1);
else
{
int x=query(a+1,b+1,1);
printf("%d\n",x);
}
}
}

return 0;
}

1. 第一句可以忽略不计了吧。从第二句开始分析，说明这个花色下的所有牌都会在其它里面出现，那么还剩下♠️和♦️。第三句，可以排除2和7，因为在两种花色里有。现在是第四句，因为♠️还剩下多个，只有是♦️B才能知道答案。

2. 很高兴你会喜欢这个网站。目前还没有一个开发团队，网站是我一个人在维护，都是用的开源系统，也没有太多需要开发的部分，主要是内容整理。非常感谢你的关注。