LCIS（线段树区间合并）

LCIS

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5912    Accepted Submission(s): 2563

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].

 

 

Input

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).

 

 

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

 

 

题解：

给你n个数组成的序列（序列中编号从0到n-1），有q次操作。

操作1——Q a  b，让你输出区间[a, b]里面最长的连续递增序列的长度；

操作2——U a  b，修改序列第a个数为b。

出了点小问题错了半天；

刚开始理解错了，是单调上升序列；1 3 5 7也算；

区间合并

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define SD(x,y) scanf("%lf%lf",&x,&y)
#define P_ printf(" ")
#define ll root<<1
#define rr root<<1|1
#define lson ll,l,mid
#define rson rr,mid+1,r
typedef long long LL;
const int MAXN=100010;
struct Node{
	int len,lv,rv,lsum,rsum,sum;
	 Node init(int l,int r){
		scanf("%d",&lv);rv=lv;
		lsum=sum=rsum=1;
	}
};
Node tree[MAXN<<2];
/*void print(int root,int l,int r){
	int mid=(l+r)>>1;
	if(l==r){
		printf("%d %d\n",tree[root].lv,tree[root].rv);return;
	}
	print(lson);print(rson);
}*/

void pushup(int root){
	tree[root].lsum=tree[ll].lsum;
	tree[root].rsum=tree[rr].rsum;
	tree[root].lv=tree[ll].lv;
	tree[root].rv=tree[rr].rv;
	tree[root].sum=max(tree[ll].sum,tree[rr].sum);
	if(tree[ll].rv<tree[rr].lv){
		if(tree[ll].lsum==tree[ll].len)tree[root].lsum+=tree[rr].lsum;
		if(tree[rr].rsum==tree[rr].len)tree[root].rsum+=tree[ll].rsum;
		tree[root].sum=max(tree[root].sum,tree[ll].rsum+tree[rr].lsum);
	}
}

void build(int root,int l,int r){
	tree[root].len=r-l+1;//放在里面了。。。错了 
	if(l==r){
		tree[root].init(l,r);
		return ;
	}
	int mid=(l+r)>>1;
	build(lson);
	build(rson);
	pushup(root);
}
void update(int root,int l,int r,int A,int B){
	int mid=(l+r)>>1;
	if(l==A&&r==A){
		tree[root].lv=tree[root].rv=B;
		return;
	}
	if(mid>=A)update(lson,A,B);
	else update(rson,A,B);
	pushup(root);
}

int query(int root,int l,int r,int A,int B){
	if(l>=A&&r<=B)return tree[root].sum;
	int mid=(l+r)>>1;
	int ans=0;
	if(mid>=A)ans=max(ans,query(lson,A,B));//多加了return错了半天；；；； 
	if(mid<B)ans=max(ans,query(rson,A,B));//
	if(tree[ll].rv<tree[rr].lv)
		ans=max(ans,min(mid-A+1,tree[ll].rsum)+min(B-mid,tree[rr].lsum));
		//以免超过查询区间；因为里面的lsum,rsum分别维护的左右最大连续上升序列； 
	return ans;
}
int main(){
	int T,N,M;
	char s[5];
	int A,B; 
	SI(T);
	while(T--){
		SI(N);SI(M);
		build(1,1,N);
		while(M--){
			scanf("%s%d%d",s,&A,&B);
			A++;
			if(s[0]=='Q'){
				B++;
				printf("%d\n",query(1,1,N,A,B));
			}
			else update(1,1,N,A,B);
		//	print(1,1,N);
		}
	} 
	return 0;
}