题目描述
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。
输入
The first line contains integer n (1 ≤ n ≤ 105), showing how many numbers the sequence has. The next line contains n integers a1, a2, ..., an (|ai| ≤ 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.
输出
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.
样例输入
9
9 -8 9 -1 -1 -1 9 -8 9
3
1 1 9 1
1 1 9 2
1 4 6 3
样例输出
17
25
0
题解
模拟费用流+线段树区间合并
一开始想了个类似于dp的线段树区间合并结果一看数据范围果断放弃了。。。看了题解才知道是模拟费用流。。。
考虑如果用费用流的话怎么处理：每个点有一个大小为点权的费用，每次选择一段区间，获得这些点权和的费用，然后反向边使得它们的费用取相反数。
这个过程需要维护最大连续子段和（及其位置）、支持区间翻转。可以使用线段树来维护。
每个节点维护这段区间的区间和，包含左端点的最大连续子段和、包含右端点的最大连续子段和、整体的最大连续子段和，以及最小连续字段和；还要维护翻转标记。同时，对于和及连续字段和还要维护出现的区间位置。
每个询问不断的找区间内最大连续子段和，如果其大于0则取出并区间取相反数（模拟增广的过程）。最后再把这些取了相反数的区间还原回来。
代码量极大。。。强烈建议使用结构体重载运算符以减少代码量。
时间复杂度$O(nk\log n)$。
#include <cstdio>
#include <algorithm>
#define N 100010
#define lson l , mid , x << 1
#define rson mid + 1 , r , x << 1 | 1
using namespace std;
struct data
{
int v , l , r;
data() {}
data(int V , int L , int R) {v = V , l = L , r = R;}
bool operator<(const data &a)const {return v < a.v;}
data operator+(const data &a)const {return data(v + a.v , l , a.r);}
data operator-()const {return data(-v , l , r);}
}now , sta[25];
struct seg
{
data vsum , lmax , lmin , rmax , rmin , tmax , tmin;
int rev;
seg() {}
seg(int v , int p)
{
vsum = data(v , p , p) , rev = 0;
if(v > 0)
{
lmax = rmax = tmax = data(v , p , p);
lmin = data(0 , p , p - 1) , rmin = data(0 , p + 1 , p) , tmin = data(0 , 0 , 0);
}
else
{
lmax = data(0 , p , p - 1) , rmax = data(0 , p + 1 , p) , tmax = data(0 , 0 , 0);
lmin = rmin = tmin = data(v , p , p);
}
}
seg operator+(const seg &a)const
{
seg ans;
ans.vsum = vsum + a.vsum;
ans.lmax = max(lmax , vsum + a.lmax) , ans.lmin = min(lmin , vsum + a.lmin);
ans.rmax = max(a.rmax , rmax + a.vsum) , ans.rmin = min(a.rmin , rmin + a.vsum);
ans.tmax = max(rmax + a.lmax , max(tmax , a.tmax)) , ans.tmin = min(rmin + a.lmin , min(tmin , a.tmin));
ans.rev = 0;
return ans;
}
}a[N << 2];
inline void pushup(int x)
{
a[x] = a[x << 1] + a[x << 1 | 1];
}
inline void rever(int x)
{
swap(a[x].lmax , a[x].lmin) , swap(a[x].rmax , a[x].rmin) , swap(a[x].tmax , a[x].tmin);
a[x].vsum = -a[x].vsum;
a[x].lmax = -a[x].lmax , a[x].lmin = -a[x].lmin;
a[x].rmax = -a[x].rmax , a[x].rmin = -a[x].rmin;
a[x].tmax = -a[x].tmax , a[x].tmin = -a[x].tmin;
a[x].rev ^= 1;
}
inline void pushdown(int x)
{
if(a[x].rev) rever(x << 1) , rever(x << 1 | 1) , a[x].rev = 0;
}
void build(int l , int r , int x)
{
if(l == r)
{
int v;
scanf("%d" , &v) , a[x] = seg(v , l);
return;
}
int mid = (l + r) >> 1;
build(lson) , build(rson);
pushup(x);
}
void modify(int p , int v , int l , int r , int x)
{
if(l == r)
{
a[x] = seg(v , l);
return;
}
pushdown(x);
int mid = (l + r) >> 1;
if(p <= mid) modify(p , v , lson);
else modify(p , v , rson);
pushup(x);
}
void update(int b , int e , int l , int r , int x)
{
if(b <= l && r <= e)
{
rever(x);
return;
}
pushdown(x);
int mid = (l + r) >> 1;
if(b <= mid) update(b , e , lson);
if(e > mid) update(b , e , rson);
pushup(x);
}
seg query(int b , int e , int l , int r , int x)
{
if(b <= l && r <= e) return a[x];
pushdown(x);
int mid = (l + r) >> 1;
if(e <= mid) return query(b , e , lson);
else if(b > mid) return query(b , e , rson);
else return query(b , e , lson) + query(b , e , rson);
}
int main()
{
int n , m , opt , x , y , z , tot , ans;
scanf("%d" , &n);
build(1 , n , 1);
scanf("%d" , &m);
while(m -- )
{
scanf("%d%d%d" , &opt , &x , &y);
if(opt == 1)
{
scanf("%d" , &z) , tot = ans = 0;
while(tot < z)
{
now = query(x , y , 1 , n , 1).tmax;
if(now.v <= 0) break;
ans += now.v , update(now.l , now.r , 1 , n , 1);
sta[++tot] = now;
}
printf("%d\n" , ans);
while(tot) update(sta[tot].l , sta[tot].r , 1 , n , 1) , tot -- ;
}
else modify(x , y , 1 , n , 1);
}
return 0;
}