topic
original
Consider integer sequence a1, a2, ..., an. You should run queries of two types:
- The query format is "0 i val". In reply to this query you should make the following assignment: ai = val.
- The query format is "1 l r k". In reply to this query you should print the maximum sum of at most k non-intersecting subsegments of sequence al, al + 1, ..., ar. Formally, you should choose at most k pairs of integers (x1, y1), (x2, y2), ..., (xt, yt) (l ≤ x1 ≤ y1 < x2 ≤ y2 < ... < xt ≤ yt ≤ r; t ≤ k) such that the sum ax1 + ax1 + 1 + ... + ay1 + ax2 + ax2 + 1 + ... + ay2 + ... + axt + axt + 1 + ... + ayt is as large as possible. Note that you should choose at most k subsegments. Particularly, you can choose 0 subsegments. In this case the described sum considered equal to zero.
Chinese
To give you a length \ (n-\) sequence, each time you give an interval \ ([L, R & lt] \) , asked to select no more than this range where \ (K \) stretch of consecutive elements, the biggest elements and how much. Support modification. Modify interrogation total \ (m \) times
\ (1 \ leqslant n, m \ leqslant 10 ^ 5 \ qquad 1 \ leqslant k \ leqslant20 \ qquad \) will not burst int
Thinking
Data structure, ideology
- Cost Flow simulation
- Segment tree
About cost flow simulation
- Is essentially a cost flow chart observed and found that it has some special properties, then use these properties accelerate augmented.
- Since most of these questions secret is heavy, you just want to be greedy absurd
True ideas
- Let me talk Conclusion: is to take \ (k \) sub-interval and the largest sub-segment, and take off after each interval corresponds to the inverse (opposite number becomes). If the maximum sub-segment and which one is negative, and it can be terminated early.
- First by greed and think about, take a look at why it is right to do so.
- During this last sequence, period must be selected, the section is not selected, the spacer come.
- Well, a certain period chosen not to continue to the edge of the expansion, there is definitely a reason - to either side of the two sides, or is a negative number.
- The selected interval for each segment, the leftmost and rightmost must be a positive number.
- So, every time we choose, be sure to be elected to two breakpoints final answer (ie a breakpoint selected local section of the border is not selected, apparently at most \ (2k \) th), and will not repeat (repeat the words previously selected is no longer the largest sub-segment and the). So we finally get up to \ (2k \) breakpoints, just will be able to get an answer.
- Reconsider cost flow
- FIG cost found stream, is connected a point S (0 Cost / flow k) to the side of point A, the selected interval each point has a representative node. A point to each node represents the point of even (this right cost point value / flow 1) side, each node again (the cost side the next point of the weight / flow 1) a node even after his a Finally, each point are then connected to the T point a (cost 0/1 flow) side.
- So our process is much like the flow and expenses. It corresponds to the inverse of the reverse side to come up.
- Then, we will support three modes of operation:
- A single point of modification
- The maximum range of queries and sub-segment
- Interval negated
- We can be complete segment tree is labeled a bit much (I wrote 17).
Code
#include<bits/stdc++.h>
#define LL long long
#define MAXN 100000
using namespace std;
template<typename T>void Read(T &cn)
{
char c;int sig = 1;
while(!isdigit(c = getchar()))if(c == '-')sig = -1; cn = c-48;
while(isdigit(c = getchar()))cn = cn*10+c-48; cn*=sig;
}
template<typename T>void Write(T cn)
{
if(cn<0) {putchar('-'); cn = 0-cn; }
int wei = 0; T cm = 0; int cx = cn%10; cn/=10;
while(cn)wei++,cm = cm*10+cn%10,cn/=10;
while(wei--)putchar(cm%10+48),cm/=10;
putchar(cx+48);
}
struct qwe{
int l,r;
};
struct dan{
int l,r,ans;
};
struct dx{
int zuo,you,da,he;
int l,r,z,y;
};
dan mk(int cn,int cm,int cx) {dan guo; guo.l = cn; guo.r = cm; guo.ans = cx; return guo; }
struct Seg{
struct node{
dx d,x;
int p;
};
node t[MAXN*4+1];
void zhuang(dx &cn,int cm,int cx)
{
cn.zuo = cn.you = cn.da = cn.he = cm;
cn.l = cn.r = cn.z = cn.y = cx;
}
void build(int cn,int l,int r,int a[])
{
t[cn].p = 0;
if(l == r){
zhuang(t[cn].d,a[l],l); zhuang(t[cn].x,-a[l],l);
return;
}
build(cn<<1,l,(l+r)>>1,a); build((cn<<1)|1,((l+r)>>1)+1,r,a);
update(cn,(l+r)>>1);
}
void r_update(dx &cn,dx ls,dx rs,int zh)
{
cn.he = ls.he + rs.he;
if(ls.da > rs.da)cn.da = ls.da,cn.l = ls.l,cn.r = ls.r;
else cn.da = rs.da,cn.l = rs.l,cn.r = rs.r;
if(cn.da < ls.you + rs.zuo)cn.da = ls.you + rs.zuo,cn.l = ls.y,cn.r = rs.z;
if(ls.zuo > ls.he + rs.zuo)cn.zuo = ls.zuo,cn.z = ls.z;
else cn.zuo = ls.he + rs.zuo,cn.z = rs.z;
if(cn.zuo < ls.he)cn.zuo = ls.he,cn.z = zh;
if(rs.you > rs.he + ls.you)cn.you = rs.you,cn.y = rs.y;
else cn.you = rs.he + ls.you,cn.y = ls.y;
if(cn.you < rs.he)cn.you = rs.he,cn.y = zh+1;
}
void update(int cn,int zh)
{
int ls = cn<<1,rs = ls+1;
r_update(t[cn].d,t[ls].d,t[rs].d,zh); r_update(t[cn].x,t[ls].x,t[rs].x,zh);
}
void tui(int cn)
{
if(!t[cn].p)return;
int ls = cn<<1,rs = (cn<<1)|1;
t[ls].p ^= 1; swap(t[ls].d,t[ls].x);
t[rs].p ^= 1; swap(t[rs].d,t[rs].x);
t[cn].p = 0;
}
void gai(int cn,int cm,int cx,int l,int r)
{
if(l == r) {
zhuang(t[cn].d,cx,l); zhuang(t[cn].x,-cx,l);
return;
}
tui(cn);
int zh = (l+r)>>1;
if(cm <= zh)gai(cn<<1,cm,cx,l,zh); else gai((cn<<1)|1,cm,cx,zh+1,r);
update(cn,(l+r)>>1);
}
void qufan(int cn,int cl,int cr,int l,int r)
{
if(cl <= l && r <= cr) {
dx lin = t[cn].d; t[cn].d = t[cn].x; t[cn].x = lin;
t[cn].p ^= 1;
return;
}
tui(cn);
int zh = (l+r)>>1;
if(cl <= zh)qufan(cn<<1,cl,cr,l,zh);
if(cr > zh)qufan((cn<<1)|1,cl,cr,zh+1,r);
update(cn,(l+r)>>1);
}
dx xun(int cn,int cl,int cr,int l,int r)
{
if(cl <= l && r <= cr)return t[cn].d;
tui(cn);
int zh = (l+r)>>1;
if(cl <= zh){
dx guo = xun(cn<<1,cl,cr,l,zh);
if(cr > zh){
dx guo2 = xun((cn<<1)|1,cl,cr,zh+1,r);
r_update(guo,guo,guo2,zh);
}
return guo;
}
if(cr > zh)return xun((cn<<1)|1,cl,cr,zh+1,r);
}
}T;
int n,q;
int a[MAXN+1];
qwe zhan[MAXN+1];
int zlen;
int main()
{
// freopen("a.in","r",stdin);
// freopen("a.ou","w",stdout);
Read(n);
for(int i = 1;i<=n;i++)Read(a[i]);
T.build(1,1,n,a);
Read(q);
for(int i = 1;i<=q;i++)
{
int bx,bn,bm,by;
Read(bx); Read(bn); Read(bm);
if(!bx)T.gai(1,bn,bm,1,n);
else{
Read(by);
zlen = 0;
LL guo = 0;
for(int j = 1;j<=by;j++)
{
dx lin = T.xun(1,bn,bm,1,n);
if(lin.da <= 0)break;
guo += lin.da;
zhan[++zlen].l = lin.l; zhan[zlen].r = lin.r;
T.qufan(1,lin.l,lin.r,1,n);
}
for(int j = 1;j<=zlen;j++)T.qufan(1,zhan[j].l,zhan[j].r,1,n);
Write(guo); putchar('\n');
}
}
return 0;
}