제출 #626886

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
626886		Monchito	메기 농장 (IOI22_fish)	C++17	53 / 100	1089 ms	18916 KiB

fish

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

#define fst first
#define snd second
#define pb push_back
#define mp make_pair
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()

typedef long long ll;
typedef vector<int> vi;

const int MAXN = 1e5 + 2;

int n, m;
vector<pair<int,ll>> a[MAXN];

ll solve1(){
  ll sum=0;
  for(int i=0; i<n; i++){
    for(auto& [y, w] : a[i]) sum += w;
  }
  return sum;
}

ll solve2(){
  ll res1=0, res2=0, res3=0;
  for(auto& [y, w] : a[0]) res1 += w;
  for(auto& [y, w] : a[1]) res2 += w;

  if(n>2){
    res3=res2;
    ll acc=res2;
    int i=0, j=0;
    for(int h=0; h<n; h++){
      while(j<sz(a[1]) && a[1][j].fst<=h) acc -= a[1][j++].snd;
      while(i<sz(a[0]) && a[0][i].fst<=h) acc += a[0][i++].snd;
      res3=max(res3,acc);
    }
  }
  return max(res1,max(res2,res3));
}

ll dp1[MAXN][2];
ll solve3(int i, int flag){
  if(i==n) return 0;
  ll& ret = dp1[i][flag];
  if(ret != -1) return ret;
  ret = solve3(i+1, 1);
  if(sz(a[i])){
    if(flag==1) ret=max(ret,a[i][0].snd+solve3(i+1,0));
    if(i<n-1) ret=max(ret,a[i][0].snd+solve3(i+2,1));
  }
  return ret;
}

ll dp2[300][1000];

ll solve4(int i, int v){
  if(i==n) return 0;
  ll& ret = dp2[i][v];
  if(ret != -1) return ret;

  int prev_used = (v <= n) ? v : 0, curr_used = (v > n) ? v - n : 0;

  int l=-1,r=-1;
  ll acc=0;
  for(int j=0; j<sz(a[i]); j++){
    if(a[i][j].fst < curr_used) continue;
    if(l==-1) l=r=j;
    if(a[i][j].fst < prev_used) acc+=a[i][j].snd, r++;
    else break;
  }

  ret = solve4(i+1, 0) + acc;
  ll curr=acc;
  for(int h=0; h<n; h++){
    while(l<r && a[i][l].fst <= h) curr-=a[i][l++].snd;
    ret=max(ret,curr+solve4(i+1, h+1));
  }
  if(i<n-1 && r!=-1){
    for(int j=r; j<sz(a[i]); j++){
      acc += a[i][j].snd;
      ret=max(ret, acc+solve4(i+1, a[i][j].fst+1+n));
    }
  }
  return ret;
}

ll max_weights(int N, int M, vi X, vi Y, vi W) {
  bool subtask1=true, subtask2=true, subtask3=true, subtask4_5=n<=300;
  n = N, m = M;
  for(int i=0; i<m; i++){
    subtask1 &= X[i] % 2 == 0;
    subtask2 &= X[i] <= 1;
    subtask3 &= Y[i] == 0;
    a[X[i]].pb(mp(Y[i], ll(W[i])));
  }
  for(int i=0; i<n; i++) sort(all(a[i]));

  if(subtask1) return solve1();
  if(subtask2) return solve2();
  if(subtask3){
    memset(dp1,-1,sizeof(dp1));
    return solve3(0, 0);
  }
  if(subtask4_5){
    memset(dp2, -1, sizeof(dp2));
    return solve4(0, 0);
  }
  return 0;
}

• Subtask 1: 3 / 3
• Subtask 2: 6 / 6
• Subtask 3: 9 / 9
• Subtask 4: 14 / 14
• Subtask 5: 21 / 21
• Subtask 6: 0 / 17
• Subtask 7: 0 / 14
• Subtask 8: 0 / 16