(Help, Silver) Closest Cow Wins

Problem Info

Closest Cow Wins


My Work

#include <iostream>
#include <fstream>
#include <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <math.h>
#include <map>
#include <set>
#include <stack>
#include <queue>

using namespace std;

long long K, M, N;

vector<pair<long long, long long>> patches;
vector<long long> nhoj;

int main()
	// Input
	cin >> K >> M >> N;
	long long t1, t2;
	for (int i = 0; i < K; i++) {
		cin >> t1 >> t2;
		patches.push_back({ t1, t2 });
	for (int i = 0; i < M; i++) {
		cin >> t1;
	// Make sure the data is sorted 
	sort(patches.begin(), patches.end(),
		[](pair<long long, long long> a, pair<long long, long long> b) {
			return a.first < b.first;
	sort(nhoj.begin(), nhoj.end());
	vector<long long> answers; // The amounts of tastiness we can get by adding a single cow 
	// Get the tastiness before the first cow of Farmer Nhoj 
	int idx = 0;
	long long before = 0;
	while (patches.at(idx).first < nhoj.front()) {
		before += patches.at(idx).second;
	// Get the maximum tastiness between each set of two of Farmer Nhoj's cows 
	// We calculate the maximum we can get from just placing one cow, and also the total (for when we want to place two cows) 
	for (int i = 0; i < nhoj.size() - 1; i++) {
		long long oneCow = 0; // Max tastiness we can get if we only place one cow in this region 
		int start = nhoj.at(i); 
		int end = nhoj.at(i + 1);
		int range = (end - start) / 2; // Range of patches when we palce one cow 
		int idx2 = idx;
		long long currWindow = 0; // Local value for oneCow with current "idx" var 
		long long total = 0; // Total patch tastiness in this region 
		while (idx < patches.size() && patches.at(idx).first < end) {
			if (patches.at(idx).first == start) continue; // Make sure we don't count one that a Nhoj cow is already sitting on with since we can never get those patches 
			total += patches.at(idx).second;
			while (idx2 < patches.size() && patches.at(idx2).first < patches.at(idx).first + range && idx2 < end) {
				currWindow += patches.at(idx2).second;
			oneCow = max(oneCow, currWindow);
			currWindow -= patches.at(idx).second;
		answers.push_back(oneCow); // Push the max tastiness we can get with one cow 
		answers.push_back(total - oneCow); // Push how much tastiness we would add if we put in two cows instead of one 
	// Get tastiness after last Nhoj cow 
	int indexThingy = patches.size() - 1;
	long long after = 0;
	while (patches.at(indexThingy).first > nhoj.back()) {
		after += patches.at(indexThingy).second;
	// Sort answers and do output 
	sort(answers.begin(), answers.end(), greater<long long>());
	long long ans = 0;
	for (int i = 0; i < N; i++) {
		ans += answers.at(i);
	cout << ans << "\n";
	return 0;

Sorry if my explanation is a little wacky. Basically, my code goes through each “bubble” between two of Nhoj’s cows. If we place two cows in each region, it is guaranteed that we can get all the patches within that bubble. If we only place one, we’ll only get part of it. My code goes through each bubble (as well as before the first Nhoj cow and after the last one) and calculates the total possible tastiness if we use two cows, as well as the max tastiness if we only use one. It pushes the amount of tastiness can be added by a cow in different locations to a vector, sorts it, adds together the first N elements, and outputs the answer.

What I’ve Tried

I’ve tried downloading the test data but it’s huge so I don’t know what’s really wrong with my code. I also tried changing the sample case input but my code behaves as expected every time.


I only pass the sample case, everything else is wrong. Can someone please explain what I’m doing wrong? I know I probably did something dumb but this is like my third attempt at implementing this and I’ve spent an embarrassingly long time on this and I’m honestly exhausted at this point. Any help is appreciated.


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