Submission #978199

#TimeUsernameProblemLanguageResultExecution timeMemory
978199asdfgraceAdvertisement 2 (JOI23_ho_t2)C++17
69 / 100
2094 ms14076 KiB
#include <bits/stdc++.h> using namespace std; #define dbg(x) x #define prt(x) dbg(cerr << x) #define pv(x) dbg(cerr << #x << " = " << x << '\n') #define pv2(x) dbg(cerr << #x << " = " << x.first << ',' << x.second << '\n') #define parr(x) dbg(prt(#x << " = { "); for (auto y : x) prt(y << ' '); prt("}\n");) #define parr2(x) dbg(prt(#x << " = { "); for (auto [y, z] : x) prt(y << ',' << z << " "); prt("}\n");) #define parr2d(x) dbg(prt(#x << ":\n"); for (auto arr : x) {parr(arr);} prt('\n')); #define parr2d2(x) dbg(prt(#x << ":\n"); for (auto arr : x) {parr2(arr);} prt('\n')); /* for any given pair it's ez to check if you can create an edge between them probably can calc contrib this way you can quickly check like how many people you can donate this book to? note that you can only donate to someone else if your e-value is greater than theirs at the very least 2^n sub is ez what about n^2 then you get a dag with each edge existing if i can infl j ans = # of nodes with indeg 0 */ int main() { ios::sync_with_stdio(0); cin.tie(0); int n; cin >> n; vector<array<int, 2>> xe(n); int mne = 1e9 + 1, mxe = -1; for (int i = 0; i < n; i++) { cin >> xe[i][0] >> xe[i][1]; mne = min(mne, xe[i][1]); mxe = max(mxe, xe[i][1]); } sort(xe.begin(), xe.end()); auto it = unique(xe.begin(), xe.end()); xe.erase(it, xe.end()); n = (int) xe.size(); if (mne == mxe) { cout << n << '\n'; return 0; } vector<int> indeg(n, 0); for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (j == i) continue; if (abs(xe[i][0] - xe[j][0]) <= xe[i][1] - xe[j][1]) { indeg[j]++; } } } int cnt = 0; for (int i = 0; i < n; i++) { if (!indeg[i]) cnt++; } cout << cnt << '\n'; }
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