BZOJ2388: 旅行规划(分块 凸包)
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2022-04-15 16:38:47
题意 "题目链接" Sol 直接挂队爷的题解了 分块题好难调啊qwq cpp include define LL long long using namespace std; const int MAXN = 1e6 + 10; const LL INF = 6e18; template inlin ......
题意
sol
直接挂队爷的题解了
分块题好难调啊qwq
#include<bits/stdc++.h> #define ll long long using namespace std; const int maxn = 1e6 + 10; const ll inf = 6e18; 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;} 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, m, block, bel[maxn], bl[maxn], br[maxn], mx; vector<int> con[maxn]; ll bg[maxn], d[maxn], a[maxn]; double slope(int x, int y) { return double (a[y] - a[x]) / (y - x); } void rebuild(int id) { vector<int> &v = con[id]; v.clear(); for(int i = bl[id]; i <= br[id]; i++) { while(v.size() > 1 && (slope(v[v.size() - 2], i) >= (slope(v[v.size() - 2], v[v.size() - 1])))) v.pop_back(); v.push_back(i); } } ll find(vector<int> &v, double k, int lef) { int l = 0, r = v.size() - 1, ans = 0; while(l <= r) { int mid = l + r >> 1; if(mid == 0 || (slope(v[mid - 1], v[mid]) > k)) ans = mid, l = mid + 1; else r = mid - 1; } return a[v[ans]] + 1ll * (v[ans] - lef + 1) * (-k); } void modify(int l, int r, int val) { for(int i = l; i <= min(r, br[bel[l]]); i++) a[i] += 1ll * val * (i - l + 1); rebuild(bel[l]); if(bel[l] != bel[r]) { for(int i = bl[bel[r]]; i <= r; i++) a[i] += 1ll * val * (i - l + 1); } for(int i = r + 1; i <= br[bel[r]]; i++) a[i] += 1ll * val * (r - l + 1); rebuild(bel[r]); for(int i = bel[l] + 1; i <= bel[r] - 1; i++) { bg[i] += 1ll * (bl[i] - l + 1) * val - val; d[i] += val; } for(int i = bel[r] + 1; i <= mx; i++) bg[i] += 1ll * val * (r - l + 1); } ll query(int l, int r) { ll ans = -inf; for(int i = l; i <= min(r, br[bel[l]]); i++) chmax(ans, bg[bel[l]] + 1ll * (i - bl[bel[l]] + 1) * d[bel[l]] + a[i]); if(bel[l] != bel[r]) { for(int i = bl[bel[r]]; i <= r; i++) chmax(ans, bg[bel[r]] + 1ll * (i - bl[bel[r]] + 1) * d[bel[r]] + a[i]); } for(int i = bel[l] + 1; i <= bel[r] - 1; i++) { chmax(ans, bg[i] + find(con[i], -d[i], bl[i])); } return ans; } signed main() { n = read(); block = sqrt(n); for(int i = 1; i <= n; i++) a[i] = read() + a[i - 1], bel[i] = (i - 1) / block + 1, chmax(mx, bel[i]); // for(int i = 1; i <= 16; i++) cout << a[i] << " "; for(int i = 1; i <= mx; i++) bl[i] = (i - 1) * block + 1, br[i] = bl[i] + block - 1, rebuild(i); m = read(); for(int i = 1; i <= m; i++) { int opt = read(); if(opt == 0) { int l = read(), r = read(), v = read(); modify(l, r, v); } else { int l = read(), r = read(); cout << query(l, r) << '\n'; } } return 0; } /* 16 51 -6867 25916 -19111 -23413 -282 7274 6888 15114 6563 18264 -11811 27336 14638 19495 -10931 1 1 9 16 */
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