제출 #1180578

#제출 시각아이디문제언어결과실행 시간메모리
1180578madamadam3Catfish Farm (IOI22_fish)C++20
6 / 100
57 ms14008 KiB
#include "fish.h" #include <bits/stdc++.h> using namespace std; using vi = vector<int>; typedef long long ll; #define sz(x) (int) (x).size() #define all(x) (x).begin(), (x).end() #define pb push_back /* Time log: - first start: 1048am - 3/100: 1100am - first stop: 1128am - second start: 337pm - - second end: - overall time used: */ /* NxN grid of cells grid[c][r] = the cell at column c row r grid[0][r] is the westernmost point, and grid[N-1][r] is the easternmost point grid[c][0] = southernmost point, grid[c][N-1] northermost point M catfishes X[i], Y[i] for i < M are column and row of fish i W[i] = weight of fish i a catfish can be caught if grid[ci][ri] is empty and grid[ci-1][ri] or grid[ci+1][ri] has a pier tile a pier extends from grid[cP][0] to grid[cP][rP] */ int N, M; vi X, Y; vector<ll> W; bool cmp(int a, int b) { // compare two fish in the same col return Y[a] < Y[b]; } ll max_weights(int _N, int _M, vi _X, vi _Y, vi _W) { N = _N; M = _M; X = _X; Y = _Y; for (int i = 0; i < M; i++) W.pb(ll(_W[i])); // subtask 2: x[i] <= 1: either use everything in col0, or col1 (as using 1 means you cant use the other) ll c1 = 0; ll pc0 = 0; ll fish[2][N]; for (int i = 0; i < N; i++) fish[0][i] = fish[1][i] = 0LL; for (int i = 0; i < M; i++) { // cout << "i: " << i << " x: " << X[i] << " y: " << Y[i] << " W: " << W[i] << "\n"; if (X[i] == 1) { c1 += W[i]; fish[1][Y[i]] = W[i]; } else if (X[i] == 0) { pc0 += W[i]; fish[0][Y[i]] = W[i]; } } if (N >= 3) { ll c0 = 0; ll ans = c1; for (int i = 0; i < N; i++) { c1 -= fish[1][i]; c0 += fish[0][i]; ans = max(ans, c0 + c1); } return max(ans, c1); } else if (N == 2) { return max(c1, pc0); } }

컴파일 시 표준 에러 (stderr) 메시지

fish.cpp: In function 'll max_weights(int, int, vi, vi, vi)':
fish.cpp:77:1: warning: control reaches end of non-void function [-Wreturn-type]
   77 | }
      | ^
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