UESTC 1425 求任意区间的LIS 线段树区间更新区间查询

九野的博客，转载请注明出处：http://blog.csdn.net/acmmmm/article/details/11976621

Description

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 theLCIS (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 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].

T <= 10;

1 <= N, Q <= 100000;

1 <= L <= R <= N;

-10000 <= S1,S2,...,SN, V <= 10000.

Output

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.

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


题意：n个数字m个询问
q l r //输出 [l,r] 区间的LIS
a l r temp // [l,r] 区间+ temp

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <queue>
#define N  201000
#define M 2000100
#define inf64 0x7ffffff
#define inf 0x7ffffff
#define ll int
#define LL(x) (x<<1)
#define RR(x) (x<<1|1)
#define MID(x,y) (x+y)>>1
using namespace std;
inline ll Min(ll a,ll b){return a>b?b:a;}
inline ll Max(ll a,ll b){return a>b?a:b;}

struct node    
{    
	int lft,rht;    
	int lmx,rmx,mx;    
	int lval,rval,add;    
	void fun(int tmp)    
	{ add+=tmp; lval+=tmp; rval+=tmp; }    
	int len(){return rht-lft+1;}    
	int mid(){return MID(lft,rht);}    
	void init(){ lmx=rmx=mx=add=0; }    
};    

int y
,n,m;    

struct Segtree    
{    
	node tree[N*4];    
	void down(int ind)    //向下更新,lazy,
	{    
		if(tree[ind].add)    
		{    
			tree[LL(ind)].fun(tree[ind].add);    
			tree[RR(ind)].fun(tree[ind].add);    
			tree[ind].add=0;    
		}    
	}    
	void up(int ind)    //向上更新
	{    
		tree[ind].lmx=tree[LL(ind)].lmx;    
		tree[ind].rmx=tree[RR(ind)].rmx;    
		tree[ind].lval=tree[LL(ind)].lval;    
		tree[ind].rval=tree[RR(ind)].rval;    
		tree[ind].mx=max(tree[LL(ind)].mx,tree[RR(ind)].mx);    

		if(tree[LL(ind)].rval<tree[RR(ind)].lval)    
		{    
			if(tree[LL(ind)].lmx==tree[LL(ind)].len())    
				tree[ind].lmx+=tree[RR(ind)].lmx;    
			if(tree[RR(ind)].rmx==tree[RR(ind)].len())    
				tree[ind].rmx+=tree[LL(ind)].rmx;    
			tree[ind].mx=max(tree[ind].mx,tree[LL(ind)].rmx+tree[RR(ind)].lmx);    
		}    
		tree[ind].mx=max(tree[ind].mx,max(tree[ind].lmx,tree[ind].rmx));    
	}    
	void build(int lft,int rht,int ind)    
	{    
		tree[ind].lft=lft;  tree[ind].rht=rht;    
		tree[ind].init();    
		if(lft==rht)    
		{    
			tree[ind].lval=tree[ind].rval= y[lft];  //把y数组更新进来  
			tree[ind].lmx=tree[ind].rmx=tree[ind].mx=1;    
		}    
		else    
		{    
			int mid=tree[ind].mid();    
			build(lft,mid,LL(ind));    
			build(mid+1,rht,RR(ind));    
			up(ind);    
		}    
	}    
	void updata(int st,int ed,int ind,int valu)    //区间更新
	{    
		int lft=tree[ind].lft,rht=tree[ind].rht;    
		if(st<=lft&&rht<=ed) tree[ind].fun(valu);    
		else    
		{    
			down(ind);    
			int mid=tree[ind].mid();    
			if(st<=mid) updata(st,ed,LL(ind),valu);    
			if(ed> mid) updata(st,ed,RR(ind),valu);    
			up(ind);    
		}    
	}    
	int query(int st,int ed,int ind)    //区间询问
	{    
		int lft=tree[ind].lft,rht=tree[ind].rht;    
		if(st<=lft&&rht<=ed) return tree[ind].mx;    
		else    
		{    
			down(ind);    
			int mid=tree[ind].mid();    
			if(ed<=mid) return query(st,ed,LL(ind));    
			else if(st>mid) return query(st,ed,RR(ind));    
			else    
			{    
				int mid=tree[ind].mid();    
				int mx1=query(st,ed,LL(ind));    
				int mx2=query(st,ed,RR(ind));    
				int tmp1=0,tmp2=0;    
				if(tree[LL(ind)].rval<tree[RR(ind)].lval)    
				{    
					tmp1=min(mid-st+1,tree[LL(ind)].rmx);    
					tmp2=min(ed-mid,tree[RR(ind)].lmx);    
				}    
				return max(max(mx1,mx2),tmp1+tmp2);    
			}    
		}    
	}    
}seg;    
//---------------线段树区间lazy更新，区间查询，LIS模版   ---------------
int main(){
	int n,a,b,i,T,Cas=1,que;
	char c;
	scanf("%d",&T);
	while(T--){
		scanf("%d %d",&n,&que);
		for(i=1;i<=n;i++)scanf("%d",&y[i]);
		seg.build( 1, n, 1);

		printf("Case #%d:\n",Cas++);
		while(que--){
			c=getchar();
			while(c!='a' && c!='q')	c=getchar();
			scanf("%d %d",&a,&b);
			if(c=='q')printf("%d\n",seg.query(a,b,1));

			else {
				scanf("%d",&i);
				seg.updata(a,b,1,i);
			}

		}
	}
	return 0;
}