bzoj3638 Cf172 k-Maximum Subsequence Sum

Posted 2020-09-20 逢山开路 遇水架桥

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=3638

【题解】

看到k<=20就感觉很py了啊

我们用一棵线段树维护选段的过程，能选到>0的段就一直选，直到选到<0的段，每次选完把段内的数全部取相反数，意为下次取是“不取”的意思。

用线段树维护左边/右边/中间的max/min

# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e5 + 10;
const int mod = 1e9+7;

# define RG register
# define ST static

int n;

struct pa {
    int l, r, x;
    pa() {}
    pa(int l, int r, int x) : l(l), r(r), x(x) {}
    friend pa operator + (pa a, pa b) {
        pa c; c.x = a.x + b.x;
        c.l = a.l, c.r = b.r;
        return c;
    }
    friend bool operator < (pa a, pa b) {
        return a.x < b.x;
    }
    friend bool operator > (pa a, pa b) {
        return a.x > b.x;
    }
};

struct querys {
    pa lmx, rmx, mx, s;
    querys() {}
    querys(pa lmx, pa rmx, pa mx, pa s) : lmx(lmx), rmx(rmx), mx(mx), s(s) {}
};

namespace SMT {
    const int Ms = 1e6 + 10;
    pa lmx[Ms], rmx[Ms], lmi[Ms], rmi[Ms], mx[Ms], mi[Ms], s[Ms];
    bool tag[Ms];        // -1
    # define ls (x<<1)
    # define rs (x<<1|1)
    inline void up(int x) {
        if(!x) return ;
        lmx[x] = max(lmx[ls], s[ls] + lmx[rs]);
        lmi[x] = min(lmi[ls], s[ls] + lmi[rs]);
        rmx[x] = max(rmx[rs], rmx[ls] + s[rs]);
        rmi[x] = min(rmi[rs], rmi[ls] + s[rs]);
        mx[x] = max(mx[ls], mx[rs]);
        mx[x] = max(mx[x], rmx[ls] + lmx[rs]);
        mi[x] = min(mi[ls], mi[rs]);
        mi[x] = min(mi[x], rmi[ls] + lmi[rs]);
        s[x] = s[ls] + s[rs];
    }
    inline void pushtag(int x) {
        if(!x) return ;
        lmx[x].x = -lmx[x].x;
        rmx[x].x = -rmx[x].x;
        lmi[x].x = -lmi[x].x;
        rmi[x].x = -rmi[x].x;
        mx[x].x = -mx[x].x;
        mi[x].x = -mi[x].x;
        s[x].x = -s[x].x;
        swap(mx[x], mi[x]);
        swap(lmx[x], lmi[x]);
        swap(rmx[x], rmi[x]);
        tag[x] ^= 1;
    }
    inline void down(int x) {
        if(!x) return ;
        if(!tag[x]) return ;
        pushtag(ls); pushtag(rs);
        tag[x] = 0;
    }
    inline void change(int x, int l, int r, int ps, int d) {
        if(l == r) {
            s[x].l = s[x].r = lmx[x].l = lmx[x].r = rmx[x].l = rmx[x].r = lmi[x].l = lmi[x].r = rmi[x].l = rmi[x].r = l;
            mx[x].l = mx[x].r = mi[x].l = mi[x].r = l;
            s[x].x = mx[x].x = mi[x].x = lmx[x].x = lmi[x].x = rmx[x].x = rmi[x].x = d;
            tag[x] = 0;
            return ;
        }
        down(x);
        int mid = l+r>>1;
        if(ps <= mid) change(ls, l, mid, ps, d);
        else change(rs, mid+1, r, ps, d);
        up(x);
    }
    
    inline void change2(int x, int l, int r, int L, int R) {
        if(L <= l && r <= R) {
            pushtag(x);
            return ;
        }
        down(x);
        int mid = l+r>>1;
        if(L <= mid) change2(ls, l, mid, L, R);
        if(R > mid) change2(rs, mid+1, r, L, R);
        up(x);
    }

    inline querys merge(querys a, querys b) {
        querys c;
        c.lmx = max(a.lmx, a.s+b.lmx);
        c.rmx = max(b.rmx, a.rmx+b.s);
        c.s = a.s + b.s;
        c.mx = max(a.mx, b.mx);
        c.mx = max(c.mx, a.rmx + b.lmx);
        return c;
    }
    
    inline querys query(int x, int l, int r, int L, int R) {
        if(L <= l && r <= R) return querys(lmx[x], rmx[x], mx[x], s[x]);
        down(x);
        int mid = l+r>>1;
        if(R <= mid) return query(ls, l, mid, L, R);
        else if(L > mid) return query(rs, mid+1, r, L, R);
        else return merge(query(ls, l, mid, L, mid), query(rs, mid+1, r, mid+1, R));
    }
    
    inline void debug(int x, int l, int r) {
        printf("x = %d, l = %d, r = %d : mx = %d, lmx = %d, rmx = %d\n", x, l, r, mx[x].x, mx[x].l, mx[x].r);
        if(l == r) return ;
        int mid = l+r>>1;
        debug(ls, l, mid);
        debug(rs, mid+1, r);
    }
}

int Left[233], Right[233], m;

int main() {
    cin >> n;
    for (int i=1, t; i<=n; ++i) {
        scanf("%d", &t);
        SMT::change(1, 1, n, i, t);
    }
    int Q, a, b, c;
    querys t;
    cin >> Q;
    while(Q--) {
        int opt; scanf("%d", &opt);
        if(!opt) {
            scanf("%d%d", &a, &b);
            SMT::change(1, 1, n, a, b);
        } else {
            scanf("%d%d%d", &a, &b, &c);
            int s = 0; m = 0;
            while(c--) {
                t = SMT::query(1, 1, n, a, b);
                if(t.mx.x < 0) break;
                else s += t.mx.x;
//                cout << t.mx.l << " " << t.mx.r << " " << t.mx.x << endl;
                SMT::change2(1, 1, n, t.mx.l, t.mx.r);
                ++m; Left[m] = t.mx.l, Right[m] = t.mx.r;
            }
            printf("%d\n", s);
            for (int i=m; i; --i) SMT::change2(1, 1, n, Left[i], Right[i]);
        }
//        SMT::debug(1, 1, n);
    }
    return 0;
}

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