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Problema Inversiones (AC) - IOI-Etapa1-Problemset15 [Optimizar Solución]
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#include <bits/stdc++.h> | |
#define X first | |
#define Y second | |
#define sz size | |
using namespace std; | |
typedef long long int lli; | |
int C,N; | |
lli Limite = 0; | |
lli AC = 0; | |
pair<int, int> A[1000002]; | |
int BIT[1000015]={0}; | |
void update(int idx, int val){ | |
while(idx<=Limite){ | |
BIT[idx]+=val; | |
idx += (idx&-idx); | |
} | |
} | |
lli query(int idx){ | |
lli sum = 0; | |
while(idx){ | |
sum += BIT[idx]; | |
idx -= (idx&-idx); | |
} | |
return sum; | |
} | |
int main(){ | |
ios_base::sync_with_stdio(0); | |
cin.tie(0); | |
int i,j,k,p; | |
cin>>C; | |
for(i=0; i<C; ++i){ | |
cin>>N; | |
AC = 0; | |
memset(BIT, 0, sizeof(BIT)); | |
for(j=1; j<=N; ++j){ | |
cin>>A[j].X>>A[j].Y; | |
if(A[j].Y>Limite){ | |
Limite = A[j].Y; | |
} | |
} | |
sort(A+1, A+N+1); | |
for(j=N; j; --j){ | |
update(A[j].Y, 1); | |
AC += query(A[j].Y-1); | |
} | |
cout<<AC<<"\n"; | |
} | |
return 0; | |
} |
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