【Codeforces】280D k-Maximum Subsequence Sum【区间内k段最大和——线段树模拟费用流】

传送门：【Codeforces】280D k-Maximum Subsequence Sum

key word：取反

my  code:

#include <stdio.h>#include <string.h>#include <vector>#include <algorithm>using namespace std ;typedef long long LL ;#define clr( a , x ) memset ( a , x , sizeof a  )#define ls ( o << 1 )#define rs ( o << 1 | 1 )#define lson ls , l , m#define rson rs , m + 1 , r#define root 1 , 1 , n#define mid ( ( l + r ) >> 1 )const int MAXN = 100005 ;struct Node {    int s , mv , lv , rv , ml , mr , l , r ;    Node () {}    Node ( int x , int v ) {        s = mv = lv = rv = v ;        ml = mr = l = r = x ;    }    Node operator + ( const Node& a ) const {        Node c ;        c.s = s + a.s ;        if ( mv > a.mv ) {            c.mv = mv ;            c.ml = ml ;            c.mr = mr ;        } else {            c.mv = a.mv ;            c.ml = a.ml ;            c.mr = a.mr ;        }        if ( rv + a.lv > c.mv ) {            c.mv = rv + a.lv ;            c.ml = r ;            c.mr = a.l ;        }        c.lv = lv ;        c.rv = a.rv ;        c.l = l ;        c.r = a.r ;        if ( s + a.lv > c.lv ) {            c.lv = s + a.lv ;            c.l = a.l ;        }        if ( rv + a.s > c.rv ) {            c.rv = rv + a.s ;            c.r = r ;        }        return c ;    }    Node operator - ( const Node& a ) const {        Node c ;        c.s = s + a.s ;        if ( mv < a.mv ) {            c.mv = mv ;            c.ml = ml ;            c.mr = mr ;        } else {            c.mv = a.mv ;            c.ml = a.ml ;            c.mr = a.mr ;        }        if ( rv + a.lv < c.mv ) {            c.mv = rv + a.lv ;            c.ml = r ;            c.mr = a.l ;        }        c.lv = lv ;        c.rv = a.rv ;        c.l = l ;        c.r = a.r ;        if ( s + a.lv < c.lv ) {            c.lv = s + a.lv ;            c.l = a.l ;        }        if ( rv + a.s < c.rv ) {            c.rv = rv + a.s ;            c.r = r ;        }        return c ;    }    void rev () {        mv = -mv ;        lv = -lv ;        rv = -rv ;        s = -s ;    }} ;struct Seg {    Node mi , ma ;    bool flip ;    void down () {        swap ( mi , ma ) ;        mi.rev () ;        ma.rev () ;        flip ^= 1 ;    }} ;Seg T[MAXN << 2] ;Node ans ;int S[30][2] ;int n , m ;void up ( int o ) {    T[o].mi = T[ls].mi - T[rs].mi ;    T[o].ma = T[ls].ma + T[rs].ma ;}void down ( int o ) {    if ( T[o].flip ) {        T[ls].down () ;        T[rs].down () ;        T[o].flip = 0 ;    }}void modify ( int x , int v , int o , int l , int r ) {    if ( l == r ) T[o].mi = Node ( x , v ) , T[o].ma = Node ( x , v ) ;    else {        down ( o ) ;        int m = mid ;        if ( x <= m ) modify ( x , v , lson ) ;        else modify ( x , v , rson ) ;        up ( o ) ;    }}void update ( int L , int R , int o , int l , int r ) {    if ( L <= l && r <= R ) T[o].down () ;    else {        down ( o ) ;        int m = mid ;        if ( L <= m ) update ( L , R , lson ) ;        if ( m <  R ) update ( L , R , rson ) ;        up ( o ) ;    }}void query ( int L , int R , int o , int l , int r ) {    if ( L <= l && r <= R ) ans = ans + T[o].ma ;    else {        down ( o ) ;        int m = mid ;        if ( L <= m ) query ( L , R , lson ) ;        if ( m <  R ) query ( L , R , rson ) ;    }}void build ( int o , int l , int r ) {    T[o].flip = 0 ;    if ( l == r ) {        int v ;        scanf ( "%d" , &v ) ;        T[o].mi = Node ( l , v ) ;        T[o].ma = Node ( l , v ) ;        return ;    }    int m = mid ;    build ( lson ) ;    build ( rson ) ;    up ( o ) ;}void solve () {    int op , x , y , v , k ;    build ( root ) ;    scanf ( "%d" , &m ) ;    for ( int i = 0 ; i < m ; ++ i ) {        scanf ( "%d" , &op ) ;        if ( op == 0 ) {            scanf ( "%d%d" , &x , &v ) ;            modify ( x , v , root ) ;        } else {            scanf ( "%d%d%d" , &x , &y , &k ) ;            int res = 0 ;            for ( int j = 0 ; j < k ; ++ j ) {                ans = Node ( -1000 , -1000 ) ;                query ( x , y , root ) ;                S[j][0] = ans.ml ;                S[j][1] = ans.mr ;                if ( ans.mv < 0 ) {                    k = j ;                    break ;                }                res += ans.mv ;                update ( S[j][0] , S[j][1] , root ) ;            }            printf ( "%d\n" , max ( res , 0 ) ) ;            for ( int j = 0 ; j < k ; ++ j ) {                update ( S[j][0] , S[j][1] , root ) ;            }        }    }}int main () {    while ( ~scanf ( "%d" , &n ) ) solve () ;    return 0 ;}