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; }
时间: 2024-10-05 20:26:58