제출 #400388

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
400388		abdzag	Cloud Computing (CEOI18_clo)	C++17	0 / 100	113 ms	262148 KiB

clo

#include<bits/stdc++.h>
#include<unordered_map>
#define rep(i,a,b) for(int i=int(a);i<int(b);i++)
#define rrep(i,a,b) for(int i=int(a);i>int(b);i--)
#define all(v) v.begin(),v.end()
#define trav(a,v) for(auto&a:v)
#define  sz(a) a.size()
typedef long double ld;
using namespace std;
const long long inf = 1e15;
typedef long long ll;
typedef unsigned long long ull;

int main()
{
    cin.sync_with_stdio(false);
    ll n,m;
    cin >> n;
    vector<pair<pair<ll, ll>, pair<ll, ll>>> v(n);
    rep(i, 0, n) {
        cin >> v[i].second.first >> v[i].first.first >> v[i].second.second;
        v[i].first.first *= -1;
    }
    cin >> m;
    v.resize(n + m);
    rep(i, n, n+m) {
        cin >> v[i].second.first >> v[i].first.first >> v[i].second.second;
        v[i].first.first *= -1;
        v[i].first.second = 1;
    }
    sort(all(v));
    vector<vector<ll>>dp(m+n+1,vector<ll>(2e5, -inf));
    dp[0][0] = 0;
    ll sumc = 1;
    rep(i, 0, n+m) {
        rep(j, 0, sumc) {
            if (!v[i].first.second) {
                if (dp[i][j] != -inf)if(v[i].second.first + j<dp[0].size())dp[i+1][v[i].second.first + j] = max(dp[i][v[i].second.first + j], dp[i][j]-v[i].second.second);
            }
            else {
                if (dp[i][j] != -inf)if (-v[i].second.first + j >= 0)dp[i+1][-v[i].second.first + j] = max(dp[i][-v[i].second.first + j], dp[i][j] + v[i].second.second);
            }
            dp[i + 1][j] = max(dp[i][j], dp[i + 1][j]);
        }
        sumc += v[i].second.first;
    }
    ll ans = 0;
    rep(i, 0, 2e5)ans = max(dp[m+n][i], ans);
    cout<<ans;
    return 0;
}

컴파일 시 표준 에러 (stderr) 메시지

clo.cpp: In function 'int main()':
clo.cpp:38:62: warning: comparison of integer expressions of different signedness: 'long long int' and 'std::vector<long long int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   38 |                 if (dp[i][j] != -inf)if(v[i].second.first + j<dp[0].size())dp[i+1][v[i].second.first + j] = max(dp[i][v[i].second.first + j], dp[i][j]-v[i].second.second);

• Subtask 1: 0 / 18
• Subtask 2: 0 / 18
• Subtask 3: 0 / 18
• Subtask 4: 0 / 18
• Subtask 5: 0 / 18
• Subtask 6: 0 / 10