我的代码(O2 AC):
#include<bits/stdc++.h>
using namespace std;
vector<int> color[55];
int prenums[200020],curprice[200020],n,p,k,premin[200020],prenode[200020];
long long ans;
int main()
{
cin>>n>>k>>p;
for(int i=1;i<=n;++i)
{
int clr;
cin>>clr>>curprice[i];
color[clr].push_back(i);
if(curprice[i]<premin[i-1]) premin[i]=curprice[i];
else premin[i]=premin[i-1];
int j=color[clr].size()-1;
/*if(curprice[i]<=p) prenums[i]=prenums[i-1]+1;
else prenums[i]=prenums[i-1];*/
}
for(int i=0;i<k;++i)
{
for(int j=0;j<color[i].size();++j)
{
for(int k=j+1;k<color[i].size();++k)
{
int nums=prenums[color[i][k]]-prenums[color[i][j]];
if(curprice[color[i][j]]<=p) nums+=1;
if(nums>0) ++ans;
}
}
}
cout<<ans<<endl;
return 0;
}
最优解:
#include <iostream>
#define maxn 200005
using namespace std;
int n,k,p;
int color,price;
int last[maxn];
int sum[maxn];
int cnt[maxn];
int ans = 0;
int now;
int main(){
cin >> n >> k >> p;
for (int i=1;i<=n;i++){
cin >> color >> price;
if (price <= p)
now = i;
if (now >= last[color])
sum[color] = cnt[color];
last[color] = i;
ans += sum[color];
cnt[color]++;
}
cout << ans << endl;
return 0;
}