이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
#define dbg(x) x
#define prt(x) dbg(cerr << x)
#define pv(x) dbg(cerr << #x << " = " << x << '\n')
#define pv2(x) dbg(cerr << #x << " = " << x.first << ',' << x.second << '\n')
#define parr(x) dbg(prt(#x << " = { "); for (auto y : x) prt(y << ' '); prt("}\n");)
#define parr2(x) dbg(prt(#x << " = { "); for (auto [y, z] : x) prt(y << ',' << z << " "); prt("}\n");)
#define parr2d(x) dbg(prt(#x << ":\n"); for (auto arr : x) {parr(arr);} prt('\n'));
#define parr2d2(x) dbg(prt(#x << ":\n"); for (auto arr : x) {parr2(arr);} prt('\n'));
/*
build 1 or 2 bridges
add to baseline of sum of all distances driven without crossing bridge + n
then only consider people crossing
k=1:
minimize the sum of all abs(s[i] - x) + abs(t[i] - x)
one of the given locs is optimal
so here treat all s[i] and t[i] the same, sort, etc.
k=2:
starts > lb and ends < rb:
lb + rb <= x + y --> use x
else --> use y
that actually applies to everything
so you can already just sort by sum
and binary search the right bound
given the left bound
but how do you compare between left bounds
okok the hint is you just sort by sum
just iterate over the n or smth breakpoints & look for the min with 2p or smth...
HINT: the median is always optimal
*/
int main() {
ios::sync_with_stdio(0); cin.tie(0);
int k, n;
cin >> k >> n;
vector<array<int, 2>> a;
long long bsl = 0;
for (int i = 0; i < n; i++) {
char c1, c2; int i1, i2;
cin >> c1 >> i1 >> c2 >> i2;
if (c1 == c2) {
bsl += abs(i2 - i1);
} else {
if (i2 < i1) swap(i1, i2);
a.push_back({i1, i2});
bsl++;
}
}
n = (int) a.size();
if (n == 0) {
cout << bsl << '\n';
return 0;
}
vector<int> b;
for (int i = 0; i < n; i++) {
b.push_back(a[i][0]); b.push_back(a[i][1]);
}
sort(b.begin(), b.end());
if (k == 1) {
long long sum = 0;
for (int i = 0; i < 2 * n; i++) {
sum += b[i] - b[0];
}
long long best = sum;
for (int i = 1; i < 2 * n; i++) {
sum += (long long) i * (b[i] - b[i - 1]);
sum -= (long long) (2 * n - i) * (b[i] - b[i - 1]);
best = min(best, sum);
}
cout << best + bsl << '\n';
} else {
sort(a.begin(), a.end(), [&] (array<int, 2> a1, array<int, 2> a2) {
return a1[0] + a1[1] < a2[0] + a2[1];
});
vector<vector<multiset<int>>> s(2, vector<multiset<int>>(2));
vector<vector<long long>> sum(2, vector<long long>(2, 0));
function<void(int, int)> ins = [&] (int i, int x) {
if (s[i][0].empty() || x <= *s[i][0].rbegin()) {
s[i][0].insert(x);
sum[i][0] += x;
if (s[i][0].size() > s[i][1].size() + 1) {
int mx = *s[i][0].rbegin();
s[i][0].erase(s[i][0].find(mx));
sum[i][0] -= mx;
s[i][1].insert(mx);
sum[i][1] += mx;
}
} else {
s[i][1].insert(x);
sum[i][1] += x;
if (s[i][1].size() > s[i][0].size()) {
int mn = *s[i][1].begin();
s[i][1].erase(s[i][1].find(mn));
sum[i][1] -= mn;
s[i][0].insert(mn);
sum[i][0] += mn;
}
}
};
function<void(int, int)> del = [&] (int i, int x) {
if (x <= *s[i][0].rbegin()) {
s[i][0].erase(s[i][0].find(x));
sum[i][0] -= x;
if (s[i][1].size() > s[i][0].size()) {
int mn = *s[i][1].begin();
s[i][1].erase(s[i][1].find(mn));
sum[i][1] -= mn;
s[i][0].insert(mn);
sum[i][0] += mn;
}
} else {
s[i][1].erase(s[i][1].find(x));
sum[i][1] -= x;
if (s[i][0].size() > s[i][1].size() + 1) {
int mx = *s[i][0].rbegin();
s[i][0].erase(s[i][0].find(mx));
sum[i][0] -= mx;
s[i][1].insert(mx);
sum[i][1] += mx;
}
}
};
function<long long(int)> cost = [&] (int i) {
if (s[i][0].empty()) return 0ll;
long long med = *s[i][0].rbegin();
return (med * (long long) s[i][0].size() - sum[i][0])
+ (sum[i][1] - med * (long long) s[i][1].size());
};
for (int i = 0; i < n; i++) {
ins(1, a[i][0]);
ins(1, a[i][1]);
}
long long best = cost(1);
for (int i = 0; i < n - 1; i++) {
del(1, a[i][0]);
del(1, a[i][1]);
ins(0, a[i][0]);
ins(0, a[i][1]);
best = min(best, cost(0) + cost(1));
}
cout << best + bsl << '\n';
}
}
/*
any observations help
check every line
IF YOUR LINES AREN'T WRONG
CHECK IF YOUR LINES ARE IN THE RIGHT ORDER
NEVER GIVE UP
*/
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