(Analysis by Nick Wu)

Due to the large number of measurements, we need to sort them in order of day.

When we process a measurement, we need to track the cow whose output is changing and the old and new outputs. After that, we need to see if the cows with maximum output have changed.

To check this, there are a few conditions we need to check. If, for example, the number of cows with the maximum output have changed, then clearly the display will need to be updated. However, the number of cows could stay the same and the display could need to still be updated. To check this, we need to investigate if the cow currently under consideration was originally up there and if they are up there after the measurement changed.

Here is Mark Gordon's code.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <set>
#include <map>

using namespace std;

struct measurement {
int day;
int cow;
int delta;
};

int main() {
ios_base::sync_with_stdio(false);
freopen("measurement.in", "r", stdin);
freopen("measurement.out", "w", stdout);

int N;
int G;
cin >> N >> G;

vector<measurement> A(N);
for (auto& m : A) {
cin >> m.day >> m.cow >> m.delta;
}
sort(A.begin(), A.end(), [](const measurement& a, const measurement& b) {
return a.day < b.day;
});

map<int, int, greater<int> > cnts;
cnts[0] = N + 1;

int result = 0;
map<int, int> mp;
for (auto& m : A) {
int& ref = mp[m.cow];

bool wastop = ref == cnts.begin()->first;
int wascnt = cnts[ref]--;
if (wascnt == 1) {
cnts.erase(ref);
}

ref += m.delta;

int iscnt = ++cnts[ref];
bool istop = ref == cnts.begin()->first;
if (wastop) {
if (!istop || wascnt != 1 || iscnt != 1) {
++result;
}
} else if (istop) {
++result;
}
}
cout << result << endl;

return 0;
}