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 pb push_back
#define mp make_pair
const int MOD=1e9+7;
struct convex{
deque<pair<long long,long long> > dq;
long long f(pair<long long,long long> line, long long x){
return line.first*x + line.second;
}
long long qry(long long x){
while(dq.size()>1){
if(f(dq[0],x)>f(dq[1],x)){ //the next line is better
dq.pop_front(); //remove useless line
}
else break;
}
return f(dq[0],x);
}
long double intersect(long long m1, long long c1, long long m2, long long c2){
return (long double)(c2-c1)/(m1-m2);
}
long double intersect(pair<long long,long long> p1, pair<long long,long long> p2){
return intersect(p1.first,p1.second,p2.first,p2.second);
}
void ins(long long m,long long c) {//insert line y=mx+c
pair<long long,long long> line = make_pair(m,c);
while (dq.size()>1) { //to prevent seg fault
long long s = dq.size();
if (intersect(dq[s-1], line) <= intersect(dq[s-2], line)){
dq.pop_back(); //removes useless line
}
else break;
}
dq.pb(line);
}
} *conv;
pair<int,int> brr[200001];
vector<pair<long long,long long> > arr;
long long take_photos(int n,int m,int k,vector<int> r,vector<int> c){
for (int i=0;i<n;i++){
r[i]++;
c[i]++;
if (r[i]>c[i]) swap(r[i],c[i]); //ri<=ci
brr[i].first=r[i];
brr[i].second=c[i];
}
sort(brr,brr+n);
arr.pb(mp(0,0));
for (int i=0;i<n;i++){
while (arr.back().first==brr[i].first){
arr.pop_back();
}
if (brr[i].second>arr.back().second){
arr.pb(brr[i]);
}
}
n=arr.size()-1;
long long dp[k+1][n+1];
dp[0][0]=0;
for (int j=1;j<=n;j++){
dp[0][j]=1e18;
}
for (int i=1;i<=k;i++){
dp[i][0]=0;
conv=new convex();
for (int j=1;j<=n;j++){
dp[i][j]=1e18;
long long grad=(1-arr[j].first)*2;
long long add=dp[i-1][j-1]+(arr[j].first-1)*(arr[j].first-1);
if (arr[j].first<=arr[j-1].second){
add-=(arr[j-1].second-arr[j].first+1)*(arr[j-1].second-arr[j].first+1);
}
conv->ins(grad,add);
dp[i][j]=conv->qry(arr[j].second);
dp[i][j]+=arr[j].second*arr[j].second;
dp[i][j]=min(dp[i][j],dp[i-1][j]);
}
}
return dp[k][n];
}
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