「BZOJ3638 / 3272」Cf172 k – Maximum Subsequence Sum

2015年4月16日5,0102

Description

给一列数，要求支持操作： 1.修改某个数的值 2.读入l,r,k，询问在[l,r]内选不相交的不超过k个子段，最大的和是多少。

Input

The first line contains integer n (1 ≤ n ≤ 10⁵), showing how many numbers the sequence has. The next line contains n integers a₁, a₂, …, a_n (|a_i| ≤ 500).

The third line contains integer m (1 ≤ m ≤ 10⁵) — the number of queries. The next m lines contain the queries in the format, given in the statement.

All changing queries fit into limits: 1 ≤ i ≤ n, |val| ≤ 500.

All queries to count the maximum sum of at most k non-intersecting subsegments fit into limits: 1 ≤ l ≤ r ≤ n, 1 ≤ k ≤ 20. It is guaranteed that the number of the queries to count the maximum sum of at most k non-intersecting subsegments doesn’t exceed 10000.

Output

For each query to count the maximum sum of at most k non-intersecting subsegments print the reply — the maximum sum. Print the answers to the queries in the order, in which the queries follow in the input.

Sample Input

9
9 -8 9 -1 -1 -1 9 -8 9
3
1 1 9 1
1 1 9 2
1 4 6 3

Sample Output

17
25
0

HINT

In the first query of the first example you can select a single pair (1, 9). So the described sum will be 17.
Look at the second query of the first example. How to choose two subsegments? (1, 3) and (7, 9)? Definitely not, the sum we could get from (1, 3) and (7, 9) is 20, against the optimal configuration (1, 7) and (9, 9) with 25.
The answer to the third query is 0, we prefer select nothing if all of the numbers in the given interval are negative.

题解

费用流构图

发现这个图比较特殊，可以手动增广

就是选择一个最大和子串，然后全部取反

线段树实现

#include<set>
#include<map>
#include<ctime>
#include<queue>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define inf 9000000000000000000LL
#define ll long long 
using namespace std;
ll read()
{
	ll x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
int n,m;
int a[100005];
vector<pair<int,int> >q;
struct data{
	int lx,rx,mx,sum;
	int lp,rp,p1,p2;
	void ini(int l,int val){
		lp=rp=p1=p2=l;
		lx=rx=mx=val;
		sum=val;
	}
};
struct seg{
	int l,r,a,b;bool flag;
	data mn,mx;
	void ini(int val){
		mn.ini(l,-val);
		mx.ini(l,val);
	}
}t[400005];
void pushdown(int k)
{
	if(t[k].l==t[k].r)return;
	if(t[k].flag)
	{
		swap(t[k<<1].mn,t[k<<1].mx);
		swap(t[k<<1|1].mn,t[k<<1|1].mx);
		t[k<<1].flag^=1;t[k<<1|1].flag^=1;t[k].flag^=1;
	}
}
data merge(data a,data b)
{
	data t;
	t.sum=a.sum+b.sum;

	t.lx=a.lx;t.lp=a.lp;
	if(a.sum+b.lx>t.lx)t.lx=a.sum+b.lx,t.lp=b.lp;

	t.rx=b.rx;t.rp=b.rp;
	if(b.sum+a.rx>t.rx)t.rx=b.sum+a.rx,t.rp=a.rp;

	t.mx=a.rx+b.lx;t.p1=a.rp;t.p2=b.lp;
	if(t.mx<a.mx)t.mx=a.mx,t.p1=a.p1,t.p2=a.p2;
	if(t.mx<b.mx)t.mx=b.mx,t.p1=b.p1,t.p2=b.p2;
	return t;
}
void update(int k)
{
	t[k].mn=merge(t[k<<1].mn,t[k<<1|1].mn);
	t[k].mx=merge(t[k<<1].mx,t[k<<1|1].mx);
}
void build(int k,int l,int r)
{
	t[k].l=l;t[k].r=r;
	if(l==r)
	{
		t[k].ini(a[l]);
		return;
	}
	int mid=(l+r)>>1;
	build(k<<1,l,mid);
	build(k<<1|1,mid+1,r);
	update(k);
}
void rever(int k,int x,int y)
{
	pushdown(k);
	int l=t[k].l,r=t[k].r,mid=(l+r)>>1;
	if(l==x&&y==r)
	{
		swap(t[k].mn,t[k].mx);t[k].flag^=1;
		return;                                                          
	}
	if(y<=mid)rever(k<<1,x,y);
	else if(x>mid)rever(k<<1|1,x,y);
	else rever(k<<1,x,mid),rever(k<<1|1,mid+1,y);
	update(k);
}
data query(int k,int x,int y)
{
	pushdown(k);
	int l=t[k].l,r=t[k].r,mid=(l+r)>>1;
	if(l==x&&y==r)return t[k].mx;
	if(y<=mid)return query(k<<1,x,y);
	if(x>mid)return query(k<<1|1,x,y);
	return merge(query(k<<1,x,mid),query(k<<1|1,mid+1,y));
}
void solve(int l,int r,int k)
{
	int ans=0;
	q.clear();
	while(k--)
	{
		data t=query(1,l,r);
		if(t.mx>0)ans+=t.mx;
		else break;
		rever(1,t.p1,t.p2);
		q.push_back(make_pair(t.p1,t.p2));
	}
	for(int i=q.size()-1;i>=0;i--)
		rever(1,q[i].first,q[i].second);
	printf("%d\n",ans);
}
void change(int k,int pos,int val)
{
	pushdown(k);
	int l=t[k].l,r=t[k].r,mid=(l+r)>>1;
	if(l==r){t[k].ini(val);return;}
	if(pos<=mid)change(k<<1,pos,val);
	else change(k<<1|1,pos,val);
	update(k);
}
int main()
{
	n=read();
	for(int i=1;i<=n;i++)a[i]=read();
	build(1,1,n);
	int opt,l,r,k;
	m=read();
	while(m--)
	{
		opt=read();l=read();r=read();
		if(opt==1){k=read();solve(l,r,k);}
		else change(1,l,r);
	}
	return 0;
}

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#include<set>

#include<map>

#include<ctime>

#include<queue>

#include<cmath>

#include<cstdio>

#include<vector>

#include<cstring>

#include<cstdlib>

#include<iostream>

#include<algorithm>

#define inf 9000000000000000000LL

#define ll long long

using namespace std;

ll read()

{

ll x=0,f=1;char ch=getchar();

while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

return x*f;

}

int n,m;

int a[100005];

vector<pair<int,int> >q;

struct data{

int lx,rx,mx,sum;

int lp,rp,p1,p2;

void ini(int l,int val){

lp=rp=p1=p2=l;

lx=rx=mx=val;

sum=val;

}

};

struct seg{

int l,r,a,b;bool flag;

data mn,mx;

void ini(int val){

mn.ini(l,-val);

mx.ini(l,val);

}

}t[400005];

void pushdown(int k)

{

if(t[k].l==t[k].r)return;

if(t[k].flag)

{

swap(t[k<<1].mn,t[k<<1].mx);

swap(t[k<<1|1].mn,t[k<<1|1].mx);

t[k<<1].flag^=1;t[k<<1|1].flag^=1;t[k].flag^=1;

}

data merge(data a,data b)

{

data t;

t.sum=a.sum+b.sum;

t.lx=a.lx;t.lp=a.lp;

if(a.sum+b.lx>t.lx)t.lx=a.sum+b.lx,t.lp=b.lp;

t.rx=b.rx;t.rp=b.rp;

if(b.sum+a.rx>t.rx)t.rx=b.sum+a.rx,t.rp=a.rp;

t.mx=a.rx+b.lx;t.p1=a.rp;t.p2=b.lp;

if(t.mx<a.mx)t.mx=a.mx,t.p1=a.p1,t.p2=a.p2;

if(t.mx<b.mx)t.mx=b.mx,t.p1=b.p1,t.p2=b.p2;

return t;

}

void update(int k)

{

t[k].mn=merge(t[k<<1].mn,t[k<<1|1].mn);

t[k].mx=merge(t[k<<1].mx,t[k<<1|1].mx);

}

void build(int k,int l,int r)

{

t[k].l=l;t[k].r=r;

if(l==r)

{

t[k].ini(a[l]);

return;

}

int mid=(l+r)>>1;

build(k<<1,l,mid);

build(k<<1|1,mid+1,r);

update(k);

}

void rever(int k,int x,int y)

{

pushdown(k);

int l=t[k].l,r=t[k].r,mid=(l+r)>>1;

if(l==x&&y==r)

{

swap(t[k].mn,t[k].mx);t[k].flag^=1;

return;

}

if(y<=mid)rever(k<<1,x,y);

else if(x>mid)rever(k<<1|1,x,y);

else rever(k<<1,x,mid),rever(k<<1|1,mid+1,y);

update(k);

}

data query(int k,int x,int y)

{

pushdown(k);

int l=t[k].l,r=t[k].r,mid=(l+r)>>1;

if(l==x&&y==r)return t[k].mx;

if(y<=mid)return query(k<<1,x,y);

if(x>mid)return query(k<<1|1,x,y);

return merge(query(k<<1,x,mid),query(k<<1|1,mid+1,y));

}

void solve(int l,int r,int k)

{

int ans=0;

q.clear();

while(k--)

{

data t=query(1,l,r);

if(t.mx>0)ans+=t.mx;

else break;

rever(1,t.p1,t.p2);

q.push_back(make_pair(t.p1,t.p2));

}

for(int i=q.size()-1;i>=0;i--)

rever(1,q[i].first,q[i].second);

printf("%d\n",ans);

}

void change(int k,int pos,int val)

{

pushdown(k);

int l=t[k].l,r=t[k].r,mid=(l+r)>>1;

if(l==r){t[k].ini(val);return;}

if(pos<=mid)change(k<<1,pos,val);

else change(k<<1|1,pos,val);

update(k);

}

int main()

{

n=read();

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

build(1,1,n);

int opt,l,r,k;

m=read();

while(m--)

{

opt=read();l=read();r=read();

if(opt==1){k=read();solve(l,r,k);}

else change(1,l,r);

}

return 0;

}

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「BZOJ3638 / 3272」Cf172 k – Maximum Subsequence Sum

Description

Input

Output

Sample Input

Sample Output

HINT

题解

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