# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
627306 | czhang2718 | Catfish Farm (IOI22_fish) | C++17 | 0 ms | 0 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
#define f first
#define s second
const int N=3e5+4;
int n,m;
vector<pair<int,int>> fish[N];
vector<ll> dp[N], dp1[N], inc[N];
long long max_weights(int n, int m, std::vector<int> X, std::vector<int> Y,
std::vector<int> W) {
for(int i=0; i<m; i++){
fish[X[i]+1].push_back({Y[i], W[i]});
}
for(int i=1; i<=n; i++){
fish[i].push_back({0, 0});
fish[i].push_back({n, 0});
sort(fish[i].begin(), fish[i].end());
int p=fish[i].size();
if(p>1 && fish[i][1].f==fish[i][0].f) fish[i].erase(fish[i].begin());
}
int i=0;
fish[i].push_back({0, 0});
dp[i].resize(2);
dp1[i].resize(2);
inc[i].resize(2);
for(int j=0; j<p; j++){
auto fi=fish[i][j];
int y=fi.f, w=fi.s;
while(nxt.f<fish[i+1].size() && fish[i+1][nxt.f].f+1<=y){
nxt.s+=fish[i+1][nxt.f].s;
nxt.f++;
}
while(prv.f<fish[i-1].size() && fish[i-1][prv.f].f+1<=y){
prv.s+=fish[i-1][prv.f].s;
prv.f++;
}
int k=upper_bound(fish[i-1].begin(), fish[i-1].end(), make_pair(y, int(1e9)))-fish[i-1].begin();
dp[i][j]=max(dp1[i-1][k]-sum, prv.s+inc[i-1][k-1]);
// cout << "inc " << prv.s+(fish[i-1][k-1].f?inc[i-1][k-1]:0) << "\n";
int k2;
if(i>=2){
k2=upper_bound(fish[i-2].begin(), fish[i-2].end(), make_pair(y, int(1e9)))-fish[i-2].begin();
dp[i][j]=max({dp[i][j], dp1[i-2][k2], dp[i-2][k2-1]+prv.s});
}
inc[i][j]=max(
j?inc[i][j-1]:ll(-1e18),
-sum + max(prv.s+inc[i-1][k-1], i>=2?max(dp1[i-2][k2], dp[i-2][k2-1]+prv.s):0));
sum+=w;
// cout << "inc[" << i << "][" << j << "] " << inc[i][j] << "\n";
// cout << "dp[" << i << "][" << j << "] " << dp[i][j] << "\n";
dp1[i][j]=dp[i][j]+nxt.s;
if(j) dp[i][j]=max(dp[i][j], dp[i][j-1]);
}
}
for(int j=p-1; j>=0; j--){
dp1[i][j]=max(dp1[i][j+1], dp1[i][j]);
}
}
return dp1[n][0];
}