제출 #1068479

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1068479		Unforgettablepl	Catfish Farm (IOI22_fish)	C++17	52 / 100	1163 ms	1840764 KiB

fish

#include "fish.h"
#include <bits/stdc++.h>
using namespace std;

long long max_weights(int N, int M,vector<int> X,vector<int> Y,
                      vector<int> W){
    vector DP(N+1,vector(N+1,vector(2,0ll)));
    vector pref(N+2,vector(N+1,0ll));
    vector transition1(N+1,vector(N+1,0ll));
    vector transition2(N+1,vector(N+1,0ll));
    vector transition3(N+1,vector(N+1,0ll));
    vector transition4(N+1,vector(N+1,0ll));
    for(int i=0;i<M;i++) {
        pref[X[i]+1][Y[i]+1]=W[i];
    }
    for(int i=1;i<=N;i++)for(int j=1;j<=N;j++)pref[i][j]+=pref[i][j-1];
    for(int i=2;i<=N;i++) {
        for(int j=0;j<=N;j++) {
            if(i!=2)DP[i][j][1]=max(DP[i][j][1],-pref[i][j]+transition1[i-1][j]);
            else {
                for(int x=j;x<=N;x++) {
                    DP[i][j][1]=max(DP[i][j][1],pref[i][x]-pref[i][j]+max(DP[i-1][x][0],DP[i-1][x][1]));
                }
            }
            if(i>2) {
                if(i!=3) {
                    DP[i][j][0]=max(DP[i][j][0],pref[i-1][j]+transition2[i-2][j]);
                    DP[i][j][0]=max(DP[i][j][0],transition3[i-2][j]);
                }
                else for(int x=0;x<=N;x++) {
                    DP[i][j][0]=max(DP[i][j][0],pref[i-1][max(x,j)]+max(DP[i-2][x][0],DP[i-2][x][1]));
                }
            }
            if(i!=2)DP[i][j][0]=max(DP[i][j][0],pref[i-1][j]+transition4[i-1][j]);
            else for(int x=0;x<=j;x++) {
                DP[i][j][0]=max(DP[i][j][0],pref[i-1][j]-pref[i-1][x]+DP[i-1][x][0]);
            }
            transition1[i][j]=pref[i+1][j]+max(DP[i][j][0],DP[i][j][1]);
            transition2[i][j]=max(DP[i][j][0],DP[i][j][1]);
            transition3[i][j]=pref[i+1][j]+max(DP[i][j][0],DP[i][j][1]);
            transition4[i][j]=-pref[i][j]+DP[i][j][0];
        }
        for(int j=N-1;j>=0;j--)transition1[i][j]=max(transition1[i][j],transition1[i][j+1]);
        for(int j=1;j<=N;j++)transition2[i][j]=max(transition2[i][j],transition2[i][j-1]);
        for(int j=N-1;j>=0;j--)transition3[i][j]=max(transition3[i][j],transition3[i][j+1]);
        for(int j=1;j<=N;j++)transition4[i][j]=max(transition4[i][j],transition4[i][j-1]);
    }
    long long ans = 0;
    for(int j=0;j<=N;j++)ans=max(ans,DP[N][j][0]);
    for(int j=0;j<=N;j++)ans=max(ans,DP[N][j][1]);
    return ans;
}

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