제출 #625786

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
625786		d4xn	메기 농장 (IOI22_fish)	C++17	29 / 100	1099 ms	231532 KiB

fish

#pragma GCC optimize ("Ofast")
//#pragma GCC target ("avx2")
#include "fish.h"
#include <bits/stdc++.h>
using namespace std;
 
//#define int long long
#define ll long long
#define vi vector<int>
#define ii pair<int, int>
#define vii vector<ii>
#define tuple2 tuple<int, int>
#define tuple3 tuple<int, int, int>
 
const int N = 1e5;
 
int n, m;
vi c[N];
unordered_map<int, int> w[N];
unordered_map<int, ll> pre[N];
map<tuple2, ll> dp[N];
//map<tuple<int, int, int>, ll> dp; // dp[i][j][k] = maximos peces que puedo conseguir
              // con las k primeras columnas donde
              // en la columna k+1 he subido i-1
              // en la columna k+2 he subido j-1
 
ll mx(int curr, int h1, int h2) {
  if (curr == -1) return 0;
  
  tuple2 t = make_tuple(h1, h2);
  auto it = dp[curr].find(t);
  if (it != dp[curr].end()) return it->second;
  
  dp[curr][t] = mx(curr-1, 0, h1);
  it = dp[curr].find(t);
 
  if (curr+1 < n) {
    for (int h : c[curr+1]) {
      h++;
      ll cnt = 0;
 
      // ver si pillo los de la derecha
      if (curr+1 < n && h > max(h1, h2)) {
        if (curr+2 == n) {
          cnt += pre[curr+1][h-1] - pre[curr+1][h1-1];
          //assert(pre[curr+1].count(h-1) && pre[curr+1].count(h1-1));
        }
        else {
          cnt += pre[curr+1][h-1] - pre[curr+1][max(h1, h2)-1];
          //assert(pre[curr+1].count(h-1) && pre[curr+1].count(max(h1, h2)-1));
        }
      }
 
      // ver si pillo los de la izquierda      
      if (curr-1 >= 0) {
        cnt += pre[curr-1][h-1];
        //assert(pre[curr-1].count(h-1));
      }
      
 
      // ver si he eliminado alguno de los que he pillado
      cnt -= pre[curr][min(h, h1)-1];
      //assert(pre[curr].count(min(h, h1)-1));
 
      it->second = max(it->second, cnt + mx(curr-1, h, h1));
    }
  }
 
  if (curr-1 >= 0) {
    for (int h : c[curr-1]) {
      h++;
      ll cnt = 0;
 
      // ver si pillo los de la derecha
      if (curr+1 < n && h > max(h1, h2)) {
        if (curr+2 == n) {
          cnt += pre[curr+1][h-1] - pre[curr+1][h1-1];
          //assert(pre[curr+1].count(h-1) && pre[curr+1].count(h1-1));
        }
        else {
          cnt += pre[curr+1][h-1] - pre[curr+1][max(h1, h2)-1];
          //assert(pre[curr+1].count(h-1) && pre[curr+1].count(max(h1, h2)-1));
        }
      }
 
      // ver si pillo los de la izquierda      
      if (curr-1 >= 0) {
        cnt += pre[curr-1][h-1];
        //assert(pre[curr-1].count(h-1));
      }
      
 
      // ver si he eliminado alguno de los que he pillado
      cnt -= pre[curr][min(h, h1)-1];
      //assert(pre[curr].count(min(h, h1)-1));
 
      it->second = max(it->second, cnt + mx(curr-1, h, h1));
    }
  }

  return it->second;
}
 
long long max_weights(int N, int M, std::vector<int> X, std::vector<int> Y,
                      std::vector<int> W) {
  n = N;
  m = M;
 
  int mxX = 0;
  for (int i = 0; i < m; i++) {
    mxX = max(mxX, X[i]);

    w[X[i]][Y[i]] = W[i];
    c[X[i]].push_back(Y[i]);
  }
 
  if (mxX <= 1) {
    ll pre0[n], pre1[n];
    for (int i = 0; i < n; i++) {
      pre0[i] = (i ? pre0[i-1] : 0);
      pre1[i] = (i ? pre1[i-1] : 0);
      if (w[0].count(i)) pre0[i] += w[0][i];
      if (w[1].count(i)) pre1[i] += w[1][i];
    }

    ll ans = max(pre0[n-1], pre1[n-1]);
    for (int i = 0; i < n; i++) {
      ans = max(ans, pre0[i] + pre1[n-1] - pre1[i]);
    }

    if (n == 2) return max(pre0[n-1], pre1[n-1]);
    if (n >= 3) return ans;
  }

  map<int, int> heights;
  for (int i = 0; i < n; i++) {
    pre[i][-1] = 0;

    for (int &h : c[i]) heights[h]++;
 
    if (i >= 3) {
      for (int &h : c[i-3]) {
        heights[h]--;
        if (!heights[h]) heights.erase(h);
      }
    }
 
    ll sum1 = 0;
    ll sum2 = 0;
    ll sum3 = 0;
    for (auto &[h, cnt] : heights) {
      auto it = w[i].find(h);
      if (it != w[i].end()) sum3 += it->second;
      pre[i][h] = sum3;
 
      if (i-1 >= 0) {
        it = w[i-1].find(h);
        if (it != w[i-1].end()) sum2 += it->second;
        pre[i-1][h] = sum2;
      }
      if (i-2 >= 0) {
        it = w[i-2].find(h);
        if (it != w[i-2].end()) sum1 += it->second;
        pre[i-2][h] = sum1;
      }
    }
  }

  return mx(n-1, 0, 0);
}

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