이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "aliens.h"
#include <bits/stdc++.h>
#define ll int
#define sz size()
#define all(x) x.begin(),x.end()
#define vll vector<ll>
#define str string
#define pb push_back
#define pof pop_front()
#define pf push_front
#define pob pop_back()
#define bc back()
using namespace std;
const ll N=4e3+5;
struct Line{
ll m,b;
ll operator () (ll x){return m * x + b;}
bool ok(Line x, Line y){
ll t1 = (b - x.b);
ll b1 = (x.m - m);
ll t2 = (x.b - y.b);
ll b2 = (y.m - x.m);
return t1 * b2 <= t2 * b1;
}
};
ll sq(ll x){
return x*x;
}
long long dp[N][N];
long long take_photos(int n, int m, int k, vector<int> r, vector<int> c) {
ll i,q;
vll len,nxt;
{//deleting unnecessary ones
vll mx(m, m), nr;
for(i=0;i<n;i++){
if(r[i] < c[i]) swap(r[i], c[i]);
mx[r[i]] = min(mx[r[i]], c[i]);
}
for(i=m-1;i>=0;i--){
if(mx[i] > i)continue;
if(nr.empty() || mx[nr.bc] > mx[i]){
nr.pb(i);
len.pb(i-mx[i]+1);
}
}
nr.pb(-1), len.pb(-1);//changing indexation
reverse(all(nr));
reverse(all(len));
r=nr;
n=r.size()-1;
k = min(k, n);
}
c.clear();
c.pb(0);
for(i=1;i<=n;i++){
nxt.pb(r[i] - len[i]);
c.pb(nxt.bc);
len[i - 1] = r[i - 1] - nxt[i - 1] ;
len[i - 1] = max(len[i - 1], 0);
}
len[0]=0;
nxt.pb(r[n]);
memset(dp, 0x3f, sizeof(dp));
dp[0][1]=0;
vector<deque<Line>> ch(k+5);
for(i=0;i<=k;i++)ch[i].pb({-2 * nxt[0], nxt[0] * nxt[0]});
for(i=1;i<=n;i++){
for(q=1;q<=min(k,i+1);q++){
//~ if(dp[i][q] >= dp[N - 1][N - 1]) continue;
auto &dq = ch[q-1];
ll x = r[i];
while(dq.sz > 1 && dq[0](x) > dq[1](x)){
dq.pof;
}
if(i>0){
dp[i][q] = r[i] * r[i] + dq[0](x);
//~ cout<<dq[0].m<<' '<<dq[0].b<<endl;
//~ cout<<i<<' '<<q<<endl;
//~ cout<<dp[i][q]<<endl;
}
ll m = (-2) * nxt[i];
ll b = nxt[i] * nxt[i] - len[i] * len[i] + dp[i][q];
Line cur = {m, b};
//~ cout<<i<<' '<<q<<endl;
//~ cout<<m<<' '<<b<<endl;
while((ll) dq.sz > 2 && dq[dq.sz - 2].ok(dq.bc, cur)) dq.pob;
dq.pb(cur);
//~ dp[i][q] = r[i]*r[i] - 2 * r[i] * nxt[j] + nxt[j] * nxt[j] - len[j] * len[j] + dp[j][q - 1];
//~ for(j=0;j<i;j++){
//~ dp[i][q] = min(dp[i][q], dp[j][q - 1] + sq(r[i] - nxt[j]) - sq(len[j]));
//~ }
}
}
long long ans=dp[n+1][0];
for(i=0;i<=k;i++) ans=min(ans,dp[n][i]);
return ans;
}
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