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;
const int NMAX = 100000 + 5;
const int KMAX = 200 + 5;
typedef long long int lint;
const lint INF = 2E18;
int N, K;
int sPart[NMAX];
lint dpBrute[2][NMAX];
int father[KMAX][NMAX];
void brute() {
dpBrute[1][0] = INF;
for (int j = 1; j <= N; ++ j)
dpBrute[1][j] = 1LL * sPart[j] * sPart[j];
for (int i = 2; i <= K; ++ i) {
for (int j = 0; j < i; ++ j)
dpBrute[i & 1][0] = INF;
for (int j = i; j <= N; ++ j) {
pair <lint, int> sol = {1LL * (sPart[j] - sPart[j - 1]) * (sPart[j] - sPart[j - 1]) + dpBrute[(i - 1) & 1][j - 1], j - 1};
for (int k = i - 1; k <= j; ++ k)
sol = min(sol, {1LL * (sPart[j] - sPart[k - 1]) * (sPart[j] - sPart[k - 1]) + dpBrute[(i - 1) & 1][k - 1], k - 1});
dpBrute[i & 1][j] = sol.first;
father[i][j] = sol.second;
}
}
}
lint dp[2][NMAX];
bool fail;
inline long double intersect(int i, int k1, int k2) {
if (sPart[k2 - 1] == sPart[k1 - 1]) {
fail = true;
return -1;
}
else {
fail = false;
return 1.0L * (1LL * sPart[k2 - 1] * sPart[k2 - 1] + dp[(i - 1) & 1][k2 - 1] - 1LL * sPart[k1 - 1] * sPart[k1 - 1] - dp[(i - 1) & 1][k1 - 1]) / (2LL * (sPart[k2 - 1] - sPart[k1 - 1]));
}
}
void solve() {
dp[1][0] = INF;
for (int j = 1; j <= N; ++ j)
dp[1][j] = 1LL * sPart[j] * sPart[j];
deque <int> dq;
for (int i = 2; i <= K; ++ i) {
dp[i & 1][0] = INF;
for (int j = 0; j < i; ++ j)
dp[i & 1][0] = INF;
dq.clear();
for (int j = i; j <= N; ++ j) {
//Add j - 1
while (dq.size() > 1) {
int k1 = dq.at(dq.size() - 2);
int k2 = dq.at(dq.size() - 1);
int k3 = j - 1;
bool decision = (intersect(i, k1, k2) >= intersect(i, k1, k3));
assert(!fail);
if (decision)
dq.pop_back();
else
break;
}
if (dq.empty())
dq.push_back(j - 1);
else {
intersect(i, dq.back(), j - 1);
if (!fail)
dq.push_back(j - 1);
}
//While first gets beaten by second
while (dq.size() > 1) {
int k1 = dq.at(0);
int k2 = dq.at(1);
lint val1 = 1LL * (sPart[j] - sPart[k1 - 1]) * (sPart[j] - sPart[k1 - 1]) + dp[(i - 1) & 1][k1 - 1];
lint val2 = 1LL * (sPart[j] - sPart[k2 - 1]) * (sPart[j] - sPart[k2 - 1]) + dp[(i - 1) & 1][k2 - 1];
if (val2 <= val1)
dq.pop_front();
else
break;
}
int k = dq.front();
dp[i & 1][j] = 1LL * (sPart[j] - sPart[k - 1]) * (sPart[j] - sPart[k - 1]) + dp[(i - 1) & 1][k - 1];
father[i][j] = k - 1;
}
}
}
int main()
{
//freopen("data.in", "r", stdin);
cin >> N >> K;
++ K;
for (int i = 1; i <= N; ++ i) {
int val;
cin >> val;
sPart[i] = val + sPart[i - 1];
}
//brute();
solve();
cout << (1LL * sPart[N] * sPart[N] - dp[K & 1][N]) / 2 << '\n';
vector <int> sol;
int where = N;
int k = K;
while (father[k][where]) {
sol.push_back(father[k][where]);
where = father[k --][where];
}
reverse(sol.begin(), sol.end());
for (int i = 0; i < sol.size(); ++ i)
cout << sol[i] << " \n"[i + 1 == sol.size()];
return 0;
}
컴파일 시 표준 에러 (stderr) 메시지
sequence.cpp: In function 'int main()':
sequence.cpp:126:23: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
for (int i = 0; i < sol.size(); ++ i)
^
sequence.cpp:127:39: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
cout << sol[i] << " \n"[i + 1 == sol.size()];
^
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