Submission #925077

#TimeUsernameProblemLanguageResultExecution timeMemory
925077myst6Bubble Sort 2 (JOI18_bubblesort2)C++14
0 / 100
14 ms1884 KiB
#include "bubblesort2.h" #include <bits/stdc++.h> using namespace std; struct Tag { int add; Tag(int _add) : add(_add) {} Tag() : add(0) {} void apply(Tag &tag) { add += tag.add; } }; struct Info { int len, sum; Info(int _len, int _sum) : len(_len), sum(_sum) {} Info() : len(0), sum(0) {} void apply(Tag &tag) { sum += tag.add * len; } Info combine(Info &other) { return {len + other.len, sum + other.sum}; } }; struct Tree { vector<Tag> tag; vector<Info> info; Tree(int size) { tag.resize(size*4); info.resize(size*4); } void build(vector<Info> &base, int x, int xl, int xr) { if (xl == xr) { info[x] = base[xl]; } else { int xm = (xl + xr) / 2; build(base, x*2, xl, xm); build(base, x*2+1, xm+1, xr); info[x] = info[x*2].combine(info[x*2+1]); } } void pushdown(int x) { info[x].apply(tag[x]); if (x*2 < (int) tag.size()) tag[x*2].apply(tag[x]); if (x*2+1 < (int) tag.size()) tag[x*2+1].apply(tag[x]); tag[x] = {}; } void update(int l, int r, int x, int xl, int xr, Tag delta) { if (l > r) return; pushdown(x); if (l == xl && r == xr) { tag[x].apply(delta); } else { int xm = (xl + xr) / 2; update(l, min(r, xm), x*2, xl, xm, delta); update(max(l, xm+1), r, x*2+1, xm+1, xr, delta); pushdown(x); pushdown(x*2); pushdown(x*2+1); info[x] = info[x*2].combine(info[x*2+1]); } } Info query(int l, int r, int x, int xl, int xr) { if (l > r) return {}; pushdown(x); if (l == xl && r == xr) { return info[x]; } else { int xm = (xl + xr) / 2; Info left = query(l, min(r, xm), x*2, xl, xm); Info right = query(max(l, xm+1), r, x*2+1, xm+1, xr); return left.combine(right); } } }; // need to only query from a number if it is smaller than everything // to the right of it // i.e. pick the rightmost item, then the next smallest and next and so on // if it is smaller than everything to the right, then i can // count the number of items above it and subtract it from how much stuff is // on the right, to get it ans // in other words, this can be done if we use a segment tree to count number below x // we can update one of these important values // now we just need to be able to identify and track how these values change // for now let's just see if i can get 60 points vector<int> countScans(vector<int> A, vector<int> X, vector<int> V) { int N = A.size(); int Q = X.size(); set<int> AV; for (int a : A) AV.insert(a); for (int v : V) AV.insert(v); map<int,int> idxmap; int T = 0; for (int av : AV) idxmap[av] = T++; Tree tree(T); vector<Info> base(T, {1, 0}); tree.build(base, 1, 0, T-1); vector<set<int>> idx(T); for (int i=0; i<N; i++) { A[i] = idxmap[A[i]]; idx[A[i]].insert(i); tree.update(0, A[i], 1, 0, T-1, {+1}); } vector<int> ans(Q); for (int i=0; i<Q; i++) { idx[A[X[i]]].erase(X[i]); tree.update(0, A[X[i]], 1, 0, T-1, {-1}); A[X[i]] = idxmap[V[i]]; tree.update(0, A[X[i]], 1, 0, T-1, {+1}); idx[A[X[i]]].insert(X[i]); int prev = 0; for (int j=0; j<T-1; j++) { if (idx[j].empty()) continue; int k = *idx[j].rbegin(); if (k < prev) continue; ans[i] = max(ans[i], tree.query(j+1, j+1, 1, 0, T-1).sum - (T-1 - k)); prev = k; } } return ans; }
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...