Submission #1199147

#	Time	Username	Problem	Language	Result	Execution time	Memory
1199147		InvMOD	Pinball (JOI14_pinball)	C++17	51 / 100	1094 ms	1640 KiB

pinball

#include<bits/stdc++.h>

using namespace std;

#define fi first
#define se second
#define pb push_back
#define eb emplace_back

#define vi vector<int>
#define pi pair<int,int>
#define sz(v) (int)(v).size()
#define all(v) (v).begin(), (v).end()
#define compact(v) (v).erase(unique(all(v)), (v).end())

template<class T> using upq = priority_queue<T, vector<T>, greater<T>>;
template<class T> int lwrbound(const vector<T>& a, const T& b, const int s = 0){return int(lower_bound(s + all(a), b) - a.begin());}
template<class T> int uprbound(const vector<T>& a, const T& b, const int s = 0){return int(upper_bound(s + all(a), b) - a.begin());}

#define FOR(i, a, b) for(int i = (a); i <= (b); i++)
#define ROF(i, a, b) for(int i = (a); i >= (b); i--)
#define sumof(x) accumulate(all(x), 0ll)
#define dbg(x) "[" << #x " = " << (x) << "]"
#define el "\n"

using ll = long long;
using ld = long double;

template<class T> bool ckmx(T& a, const T b){return (a < b ? a = b, true : false);}
template<class T> bool ckmn(T& a, const T b){return (a > b ? a = b, true : false);}

const int N = 2e5 + 5;
const int MOD = 1e9 + 7;
const ll INF = numeric_limits<ll>::max() / 8;


/*
    imagine we are at c[i]

    [L[i], c[i]] and [c[i], R[i]] will become a "ladder"
    so now we want to go out as far as possible from c[i] in 2 direction

    minL[i], maxR[i], dp[i](0, 1)

    the condition for j to be the next ladder for i is L[i] <= c[j] <= R[i] (j < i)
*/

int n, m, L[N], R[N], c[N];

ll dp[N][2], minL[N], maxR[N], W[N];

void Main()
{
    cin >> n >> m;

    for(int i = 1; i <= n; i++){
        cin >> L[i] >> R[i] >> c[i] >> W[i];
    }

    ll ans = INF;
    for(int i = 1; i <= n; i++){
        minL[i] = L[i], maxR[i] = R[i], dp[i][0] = dp[i][1] = W[i];
        for(int j = i - 1; j >= 1; j--){
            if(L[i] <= c[j] && c[j] <= R[i]){
                if(ckmn(minL[i], minL[j])){
                    dp[i][0] = dp[j][0] + W[i];
                }
                else if(minL[i] == minL[j]) ckmn(dp[i][0], dp[j][0] + W[i]);

                if(ckmx(maxR[i], maxR[j])){
                    dp[i][1] = dp[j][1] + W[i];
                }
                else if(maxR[i] == maxR[j]) ckmn(dp[i][1], dp[j][1] + W[i]);
            }
        }

        if(minL[i] == 1 && maxR[i] == m){
            ans = min(ans, dp[i][0] + dp[i][1] - W[i]);
        }
    }

    cout << (ans == INF ? -1 : ans) << el;
}

int32_t main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    #define name "InvMOD"
    if(fopen(name".INP", "r")){
        freopen(name".INP", "r", stdin);
        freopen(name".OUT", "w", stdout);
    }

    int t = 1; while(t--) Main();
    return 0;
}

Compilation message (stderr)

pinball.cpp: In function 'int32_t main()':
pinball.cpp:92:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   92 |         freopen(name".INP", "r", stdin);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~
pinball.cpp:93:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   93 |         freopen(name".OUT", "w", stdout);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 11 / 11
• Subtask 2: 18 / 18
• Subtask 3: 22 / 22
• Subtask 4: 0 / 49