Created
October 14, 2018 04:46
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AGC028. Problem C
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#include <bits/stdc++.h> | |
#define sz(x) ((int)x.size()) | |
#define all(x) (x).begin(), (x).end() | |
using namespace std; | |
typedef long long ll; | |
typedef long double ld; | |
int n, A[100010], B[100010], P[100010][2]; | |
ll s[200010]; | |
vector<pair<int,int> > v; | |
int main() { | |
memset(P, -1, sizeof(P)); | |
scanf("%d", &n); | |
ll r1 = 0, r2 = 0; | |
for(int i=1; i<=n; i++) { | |
scanf("%d%d", A+i, B+i); | |
r1 += A[i]; r2 += B[i]; | |
v.push_back({A[i], i}); | |
v.push_back({B[i], i}); | |
} | |
sort(all(v)); | |
for(int i=0; i<sz(v); i++) { | |
s[i] += v[i].first; | |
s[i+1] += s[i]; | |
if(P[v[i].second][0] == -1) P[v[i].second][0] = i; | |
else P[v[i].second][1] = i; | |
} | |
ll r = min(r1, r2); | |
for(int i=1; i<=n; i++) { | |
ll val = 0; | |
if(n == 2); | |
else if(n-3 < P[i][0]) val = s[n-3]; | |
else if(n-3 >= P[i][1]) val = s[n-1] - A[i] - B[i]; | |
else if(n-2 == P[i][1]) val = s[n-1] - A[i] - B[i]; | |
else val = s[n-2] - v[P[i][0]].first; | |
r = min(r, val + A[i] + B[i]); | |
} | |
printf("%lld", r); | |
return 0; | |
} |
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