题目链接 Problem D
这个题类似 SPOJ GSS3
做过那个题之后其实就可以秒掉这题了。
考虑当前线段树维护的结点
在那道题的基础上,这个题要多维护几个东西,大概就是左端点的奇偶性,右端点的奇偶性。
以及当前结点代表的区间是否是一个有效的子序列。
时间复杂度$O(nlogn)$
#include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i <= (b); ++i) #define dec(i, a, b) for (int i(a); i >= (b); --i) #define ls i << 1 #define rs i << 1 | 1 #define mid ((l + r) >> 1) #define lson ls, l, mid #define rson rs, mid + 1, r typedef long long LL; const int N = 1e5 + 10; struct node{ LL c, lc, rc, ret; int lo, ro, flag; } t[N << 2]; int n, m; void pushup(int i){ t[i].c = t[ls].c + t[rs].c; t[i].ret = max(t[ls].ret, t[rs].ret); t[i].flag = 0; t[i].lc = t[ls].lc; t[i].rc = t[rs].rc; t[i].lo = t[ls].lo; t[i].ro = t[rs].ro; if (t[ls].ro ^ t[rs].lo){ t[i].ret = max(t[i].ret, t[ls].rc + t[rs].lc); if (t[ls].flag) t[i].lc = max(t[i].lc, t[ls].c + t[rs].lc); if (t[rs].flag) t[i].rc = max(t[i].rc, t[rs].c + t[ls].rc); if (t[ls].flag && t[rs].flag) t[i].flag = 1; } } void build(int i, int l, int r){ if (l == r){ scanf("%lld", &t[i].ret); t[i].c = t[i].lc = t[i].rc = t[i].ret; t[i].lo = t[i].ro = (t[i].ret & 1); t[i].flag = 1; return; } build(lson); build(rson); pushup(i); } void update(int i, int l, int r, int x, LL val){ if (l == x && l == r){ t[i].ret = t[i].c = t[i].lc = t[i].rc = val; t[i].lo = t[i].ro = (val & 1); t[i].flag = 1; return; } if (x <= mid) update(lson, x, val); else update(rson, x, val); pushup(i); } node query(int i, int l, int r, int L, int R){ if (L <= l && r <= R) return t[i]; if (R <= mid) return query(lson, L, R); if (L > mid) return query(rson, L, R); node ta = query(lson, L, mid); node tb = query(rson, mid + 1, R); node ans; ans.c = ta.c + tb.c; ans.ret = max(ta.ret, tb.ret); ans.flag = 0; ans.lc = ta.lc; ans.rc = tb.rc; ans.lo = ta.lo; ans.ro = tb.ro; if (ta.ro ^ tb.lo){ ans.ret = max(ans.ret, ta.rc + tb.lc); if (ta.flag) ans.lc = max(ans.lc, ta.c + tb.lc); if (tb.flag) ans.rc = max(ans.rc, tb.c + ta.rc); if (ta.flag && tb.flag) ans.flag = 1; } return ans; } int main(){ scanf("%d%d", &n, &m); build(1, 1, n); while (m--){ int op; scanf("%d", &op); if (op == 0){ int x, y; scanf("%d%d", &x, &y); node ans = query(1, 1, n, x, y); printf("%lld\n", ans.ret); } else{ int x; LL y; scanf("%d%lld", &x, &y); update(1, 1, n, x, y); } } return 0; }