HDOJ 5306 Gorgeous Sequence 线段树

http://www.shuizilong.com/house/archives/hdu-5306-gorgeous-sequence/

Gorgeous Sequence

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 440    Accepted Submission(s): 87

Problem Description

There is a sequence a of
length n.
We use ai to
denote the i-th
element in this sequence. You should do the following three types of operations to this sequence.



0 x y t:
For every x≤i≤y,
we use min(ai,t) to
replace the original ai's
value.

1 x y:
Print the maximum value of ai that x≤i≤y.

2 x y:
Print the sum of ai that x≤i≤y.

 

Input

The first line of the input is a single integer T,
indicating the number of testcases. 



The first line contains two integers n and m denoting
the length of the sequence and the number of operations.



The second line contains n separated
integers a1,…,an (∀1≤i≤n,0≤ai<231).



Each of the following m lines
represents one operation (1≤x≤y≤n,0≤t<231).



It is guaranteed that T=100, ∑n≤1000000, ∑m≤1000000.

 

Output

For every operation of type 1 or 2,
print one line containing the answer to the corresponding query.

 

Sample Input

1
5 5
1 2 3 4 5
1 1 5
2 1 5
0 3 5 3
1 1 5
2 1 5

 

Sample Output

5
15
3
12

Hint

Please use efficient IO method

 

Source

field=problem&key=2015+Multi-University+Training+Contest+2&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">2015 Multi-University Training Contest 2

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年07月27日 星期一 08时13分39秒
File Name     :HDOJ5306.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;

const int maxn=1000030;

int n,m;
int a[maxn];

struct Node
{
	LL ss;
	int mx,tag,cv;

	void toString() {
		printf("ss: %lld mx: %d tag: %d cv: %d\n",ss,mx,tag,cv);
	}

}T[maxn<<2];

#define lrt (rt<<1)
#define rrt (rt<<1|1)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

void push_up(int rt)
{
	T[rt].ss=T[lrt].ss+T[rrt].ss;
	T[rt].mx=max(T[lrt].mx,T[rrt].mx);
	T[rt].cv=T[lrt].cv+T[rrt].cv;
}

void pnc(int t,int l,int r,int rt)
{
	if(T[rt].tag!=0&&T[rt].tag<=t) return ;

	int all=r-l+1;
	if(T[rt].cv!=all)
	{
		T[rt].ss+=(LL)t*(all-T[rt].cv);
		T[rt].tag=T[rt].mx=t;
		T[rt].cv=all;
	}
}

void push_down(int l,int r,int rt)
{
	if(T[rt].tag)
	{
		int m=(l+r)/2;
		pnc(T[rt].tag,lson); pnc(T[rt].tag,rson);
	}
}

/// 清除掉全部大于t的标记
void fix(int t,int l,int r,int rt)
{
	if(T[rt].mx<t) return ;

	if(T[rt].tag>=t) T[rt].tag=0;
	if(l==r)
	{
		T[rt].ss=T[rt].mx=T[rt].tag;
		T[rt].cv=T[rt].tag!=0;
	}
	else
	{
		push_down(l,r,rt);
		int m=(l+r)/2;
		fix(t,lson); fix(t,rson);
		push_up(rt);
	}
}

void build(int l,int r,int rt)
{
	if(l==r)
	{
		T[rt].ss=T[rt].mx=T[rt].tag=a[l];
		T[rt].cv=1;
		return ;
	}

	T[rt].tag=0;
	int m=(l+r)/2;
	build(lson); build(rson);
	push_up(rt);
}

/// 0
void update(int L,int R,int t,int l,int r,int rt)
{
	if(T[rt].mx<=t) return ;

	if(L<=l&&r<=R)
	{
		fix(t,l,r,rt);
		if(l==r)
		{
			T[rt].ss=T[rt].mx=T[rt].tag=t;
			T[rt].cv=1;
		}
		else push_up(rt);

		pnc(t,l,r,rt);
	}
	else
	{
		push_down(l,r,rt);
		int m=(l+r)/2;
		if(L<=m) update(L,R,t,lson);
		if(R>m) update(L,R,t,rson);
		push_up(rt);
	}
}

/// 1
int query_max(int L,int R,int l,int r,int rt)
{
	if(L<=l&&r<=R) return T[rt].mx;

	push_down(l,r,rt);

	int m=(l+r)/2;
	int ret=0;

	if(L<=m) ret=max(ret,query_max(L,R,lson));
	if(R>m) ret=max(ret,query_max(L,R,rson));

	return ret;
}

/// 2
LL query_sum(int L,int R,int l,int r,int rt)
{
	if(L<=l&&r<=R) return T[rt].ss;

	push_down(l,r,rt);

	int m=(l+r)/2;

	LL ret=0;
	if(L<=m) ret+=query_sum(L,R,lson);
	if(R>m) ret+=query_sum(L,R,rson);

	return ret;
}

void show(int l,int r,int rt)
{
	printf("rt: %d %d <---> %d\n   ",rt,l,r);
	T[rt].toString();

	if(l==r) return ;

	int m=(l+r)/2;
	show(lson); show(rson);
}

/********** fast read **************/

char *ch,buf[40*1024000+5];

void nextInt(int& x)
{
	x=0;
	for(++ch;*ch<=32;++ch);
	for(x=0;'0'<=*ch;ch++) x=x*10+*ch-'0';
}

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	ch=buf-1;
	fread(buf,1,1000*35*1024,stdin);

	int T_T;
	nextInt(T_T);

	while(T_T--)
	{
		nextInt(n); nextInt(m);

		for(int i=1;i<=n;i++) nextInt(a[i]);

		build(1,n,1);

		int k,l,r,t;
		while(m--)
		{
			nextInt(k);
			if(k==0)
			{
				nextInt(l); nextInt(r); nextInt(t);
				update(l,r,t,1,n,1);
			}
			else if(k==1)
			{
				nextInt(l); nextInt(r);
				printf("%d\n",query_max(l,r,1,n,1));
			}
			else if(k==2)
			{
				nextInt(l); nextInt(r);
				//printf("%lld\n",query_sum(l,r,1,n,1));
				printf("%I64d\n",query_sum(l,r,1,n,1));
			}
		}
	}
    return 0;
}