B4085 [语言月赛 202412] 顽强拼搏奖的四种发法

#include<bits/stdc++.h>
using namespace std;
int a[110][10],b[110][110],n,t,p,k;
int ti,pi,si;
int main(){
	cin>>n>>t>>p>>k;
	for(int i=1;i<=n;i++){
		cin>>t1>>pi>>si;
		if(si==1){
			a[ti][1]=i;
			if(b[ti][pi]==0){
				a[ti][2]=i;
				a[ti][3]++;
			}
			if(a[ti][3]==1){
				a[ti][4]=i;
			}
			b[ti][pi]=1;
		}
	}
	int ac1=-1,ac2=-1,ac3=-1,ac4=-1;
	int ma1=0,ma2=0,ma3=0,ma4=0;
	for(int i=1;i<=t;i++){
		if(a[i][1]>ma1){
			ma1=a[i][1;
			ac1=i;
		}
		if(a[i][2]>ma2){
			ma2=a[i][2];
			ac2++;
		}
		if(a[i][3]<k&&a[i][2]>ma3){
			ma3=a[i][2];
			ac3=i;
		}
		if(a[i][4]>ma4){
			ma4=a[i][4];
			ac4=i;
		}
	}
	cout<<ac1<<" "<<ac2<<" "<<ac3<<" "<<ac4;
	return 0;
}