cf1136E. Nastya Hasn't Written a Legend(二分 线段树)
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2022-04-08 19:25:21
题意 "题目链接" Sol yy出了一个暴躁线段树的做法。 因为题目保证了 $a_i + k_i define Pair pair define MP(x, y) make_pair(x, y) define fi first define se second define int long lon ......
题意
sol
yy出了一个暴躁线段树的做法。
因为题目保证了
\(a_i + k_i <= a_{i+1}\)
那么我们每次修改时只需要考虑取max就行了。
显然从一个位置开始能影响到的位置是单调的,而且这些位置的每个改变量都是\((a_i + x) + \sum_{t=i}^{j-1} k_t\)
那么可以建两棵线段树分别维护这两部分的值
每次修改的时候二分出要修改的位置。
打cf一定要记得开数据结构题啊qwq
#include<bits/stdc++.h> #define pair pair<int, int> #define mp(x, y) make_pair(x, y) #define fi first #define se second #define int long long #define ll long long #define ull unsigned long long #define fin(x) {freopen(#x".in","r",stdin);} #define fout(x) {freopen(#x".out","w",stdout);} using namespace std; const int maxn = 1e6 + 10, mod = 1e9 + 7, inf = 1e9 + 10; const double eps = 1e-9; template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;} template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;} template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;} template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);} template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;} template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;} template <typename a> inline void debug(a a){cout << a << '\n';} template <typename a> inline ll sqr(a x){return 1ll * x * x;} template <typename a, typename b> inline ll fp(a a, b p, int md = mod) {int b = 1;while(p) {if(p & 1) b = mul(b, a);a = mul(a, a); p >>= 1;}return b;} inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int n, a[maxn], k[maxn], s[maxn], val[maxn]; //第一棵线段树支持区间赋值 / 区间求和 //第二棵线段树支持区间赋值 / 区间求和 ..... struct seg { #define ls k << 1 #define rs k << 1 | 1 int sum[maxn], tag[maxn], siz[maxn]; void update(int k) { sum[k] = sum[ls] + sum[rs]; } void ps(int k, int f) { sum[k] = siz[k] * f; tag[k] = f; } void pushdown(int k) { if(tag[k] == -inf) return ; ps(ls, tag[k]); ps(rs, tag[k]); tag[k] = -inf; } void build(int k, int l, int r) { siz[k] = r - l + 1; tag[k] = -inf; if(l == r) {sum[k] = val[l]; return ;} int mid = l + r >> 1; build(ls, l, mid); build(rs, mid + 1, r); update(k); } void intmem(int k, int l, int r, int ql, int qr, int v) { if(ql <= l && r <= qr) {ps(k, v); return ;} int mid = l + r >> 1; pushdown(k); if(ql <= mid) intmem(ls, l, mid, ql, qr, v); if(qr > mid) intmem(rs, mid + 1, r, ql, qr, v); update(k); } int intquery(int k, int l, int r, int ql, int qr) { if(ql <= l && r <= qr) return sum[k]; int mid = l + r >> 1; pushdown(k); if(ql > mid) return intquery(rs, mid + 1, r, ql, qr); else if(qr <= mid) return intquery(ls, l, mid, ql, qr); else return intquery(ls, l, mid, ql, qr) + intquery(rs, mid + 1, r, ql, qr); } }t[2]; int geta(int x) { return t[0].intquery(1, 1, n, x, x) + k[x] - t[1].intquery(1, 1, n, x, x); } void modify(int x, int v) { int l = x, r = n, ans = l, tmp = geta(x); while(l <= r) { int mid = l + r >> 1; if(tmp + v + k[mid] - k[x] >= geta(mid)) l = mid + 1, ans = mid; else r = mid - 1; } t[0].intmem(1, 1, n, x, ans, tmp + v); t[1].intmem(1, 1, n, x, ans, k[x]); } signed main() { n = read(); for(int i = 1; i <= n; i++) a[i] = read(); for(int i = 2; i <= n; i++) k[i] = read() + k[i - 1]; memcpy(val, a, sizeof(a)); t[0].build(1, 1, n); memcpy(val, k, sizeof(k)); t[1].build(1, 1, n); for(int i = 1; i <= n; i++) s[i] = s[i - 1] + k[i]; int q = read(); while(q--) { char c = 'g'; while(c != 's' && c != '+') c = getchar(); if(c == '+') { int x = read(), v = read(); modify(x, v); } else { int l = read(), r = read(); int pre = t[0].intquery(1, 1, n, l, r); int nxt = s[r] - s[l - 1] - t[1].intquery(1, 1, n, l, r); cout << pre + nxt << '\n'; } c = 'g'; } return 0; }
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