Submission #1291687

#TimeUsernameProblemLanguageResultExecution timeMemory
1291687enzy카니발 티켓 (IOI20_tickets)C++20
27 / 100
279 ms78116 KiB
#include "tickets.h"
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1510;
const ll inf=1e14;
int ma[maxn], me[maxn];
int v[maxn][maxn];
ll dp[maxn][maxn];
vector<int>grande[maxn], pequeno[maxn];
ll solve2(int n, int m){
	vector<vector<int>>ans(n,vector<int>(m,-1));
	for(int i=0;i<=n;i++)
		for(int j=0;j<=n;j++) dp[i][j]=-inf;
	dp[0][0]=0;
	for(int i=1;i<=n;i++){
		for(int j=0;j<=n;j++){
			int k=i-j;
			if(j) dp[j][k]=max(dp[j][k],dp[j-1][k]-me[i]);
			if(k) dp[j][k]=max(dp[j][k],dp[j][k-1]+ma[i]);
		}
	}
	// for(int i=0;i<=n;i++){
	// 	for(int j=0;j<=n;j++) cout << dp[i][j] << " ";
	// 	cout << endl;
	// }
	int a=n/2, b=n/2;
	while(a+b){
		if(!a){ // tenho q pegar o maior
			ans[a+b-1][m-1]=0;
			b--;
		}else if(!b){ // tenho q pegar o menor
			ans[a+b-1][0]=0;
			a--;
		}else if(dp[a-1][b]-me[a+b]==dp[a][b]){
			ans[a+b-1][0]=0;
			a--;
		}else{
			ans[a+b-1][m-1]=0;
			b--;
		}
	}
	allocate_tickets(ans);
	return dp[n/2][n/2];
}
ll solve4(int n, int m){
	vector<vector<int>>ans(n,vector<int>(m,-1));
	vector<pair<int,pair<int,int>>>process;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++) process.push_back({v[i][j],{i-1,j-1}});
	sort(process.begin(),process.end());
	int tot=n*m;
	ll resp=0;
	for(int i=0;i<(int)process.size();i++){
		pair<int,int>p=process[i].second;
		if(i>=tot/2){
			grande[p.first].push_back(p.second);
			resp+=process[i].first;
		}else{
			pequeno[p.first].push_back(p.second);
			resp-=process[i].first;
		}
	}
	for(int i=0;i<m;i++){
		vector<pair<int,int>>big;
		for(int j=1;j<=n;j++) big.push_back({grande[j].size(),j});
		sort(big.begin(),big.end());
		for(int j=0;j<n;j++){
			int id=big[j].second;
			if(j>=n/2){
				ans[id][pequeno[id].back()]=i;
				pequeno[id].pop_back();
			}else{
				ans[id][grande[id].back()]=i;
				grande[id].pop_back();
			}
		}
	}
	allocate_tickets(ans);
	return resp;
}
ll find_maximum(int k, vector<vector<int>> x) {
	int n=x.size(), m=x[0].size();
	for(int i=1;i<=n;i++){
		me[i]=x[i-1][0];
		ma[i]=x[i-1][m-1];
		for(int j=1;j<=m;j++) v[i][j]=x[i-1][j-1];
	}
	if(k==1) return solve2(n,m);
	else return solve4(n,m);
}
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