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:
if l <= x or y <= r, dist = r - l
else, dist = r - l + 2 * (dist btwn either end & closest bridge)
assume x < y
5 cases
1) left of x
2) contains x
3) btwn x, y
4) contains y
5) right of y
best y for each x?
everything to left/containing x is handled (i.e. everything with l <= x)
n^2 sol:
for each lb
consider the rb
and update as we go maybe with set or smth
probably maintain 3 sets or smth?
like 1 set of things ending on the left of x or containing x
these don't change
everything else ends after x
as y moves to the right
sets of things that start after x
1) starts on the right of y/containing y --> automatically use y
2) x on left and y on right --> use y if r is closer to y tha x is to l
can maintain 2 sets
when y moves to the right of a segment that used to contain it
(sort other set by rb and pop elems)
move it to the "y closer" set
probs for each elem
binary search where you would update that
and don't even maintain a set for case 2
*/
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 {
long long best = 1e18;
for (int x = 0; x < 2 * n - 1; x++) {
long long sum = 0; int c = 0;
vector<long long> dif(2 * n, 0); vector<int> cd(2 * n, 0);
for (int i = 0; i < n; i++) {
if (a[i][0] <= b[x]) {
sum += abs(a[i][0] - b[x]) + abs(a[i][1] - b[x]);
} else {
sum += abs(a[i][0] - b[x]) + abs(a[i][1] - b[x]);
c--;
int i1 = lower_bound(b.begin(), b.end(), a[i][0]) - b.begin();
dif[i1] += (a[i][1] - a[i][0]) - (a[i][1] + a[i][0] - 2 * b[i1]);
cd[i1]++;
int i2 = lower_bound(b.begin(), b.end(), a[i][1]) - b.begin();
dif[i2] += (2 * b[i2] - (a[i][0] + a[i][1])) - (a[i][1] - a[i][0]);
cd[i2]++;
if (b[2 * n - 1] >= a[i][1] + (a[i][0] - b[x])) {
int i3 = lower_bound(b.begin(), b.end(), a[i][1] + (a[i][0] - b[x])) - b.begin();
dif[i3] += ((a[i][0] + a[i][1]) - 2 * b[x]) - (2 * b[i3] - (a[i][0] + a[i][1]));
cd[i3]--;
}
}
}
for (int y = x + 1; y < 2 * n; y++) {
sum += 2ll * c * (b[y] - b[y - 1]);
c += cd[y];
sum += dif[y];
best = min(best, bsl + sum);
}
}
cout << best << '\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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