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E. Best Pair
2022-07-06 04:10:00 【whitewall_ nine】
https://codeforces.com/contest/1637/problem/E
#include<bits/stdc++.h>
using namespace std;
#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fir first
#define pb push_back
#define sec second
#define sortall(x) sort((x).begin(),(x).end())
inline int read () {
int x = 0, f = 0;
char ch = getchar();
while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
return f?-x:x;
}
template<typename T> void print(T x) {
if (x < 0) putchar('-'), x = -x;
if (x >= 10) print(x/10);
putchar(x % 10 + '0');
}
#define int long long
const int N = 1e4 + 10;
void solve() {
map<int, int> cnt;
int n, m;
cin >> n >> m;
vector<int> a(n);
for (auto &tt : a) {
cin >> tt;
cnt[tt] ++;
}
vector<pii> bad(m * 2);
rep(i, 1, m) {
int x, y;
cin>> x >> y;
bad.pb({x, y});
bad.pb({y, x});
}
sort(bad.begin(), bad.end());
vector<vector<int>> occ(n);
for (auto &[num, ct] : cnt) {
occ[ct].pb(num);
}
for (auto &t : occ)
reverse(t.begin(), t.end());
int ans = 0;
for (int cx = 1; cx < n; cx ++) {
for (auto x : occ[cx]) {
for (int cy = 1; cy <= cx; cy ++) {
for (auto y : occ[cy]) {
if (y != x && !binary_search(bad.begin(), bad.end(), pair<int, int>{x, y})) {
ans = max (ans, (cx + cy) * (x + y));
break;
}
}
}
}
}
cout << ans << endl;
}
signed main () {
int t;
cin >> t;
while (t --) solve();
}
I didn't think about this question before. I was thinking about what simple ideas can help me find such an answer . Actually, this question is right stl The use of , Because it is to find the maximum , So we invert all the numbers by enumerating the number , It can reduce the complexity , If you enumerate , First calculate the number , Then enumerate the enumeration times , Replace enumeration object
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