제출 #940567

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
940567		Macker	Restore Array (RMI19_restore)	C++14	100 / 100	87 ms	1128 KiB

restore

#include <iostream>  //cin , cout
#include <fstream>   //fin, fout
#include <stdio.h>   // scanf , pringf
#include <cstdio>
#include <algorithm> // sort , stuff
#include <stack>     // stacks
#include <queue>     // queues
#include <map>
#include <string>
#include <string.h>
#include <set>
#include <assert.h>     /* assert */


using namespace std;

typedef pair<int, int>          pii;
typedef vector<int>             vi;     /// adjlist without weight
typedef vector<pii>             vii;    /// adjlist with weight
typedef vector<pair<int,pii>>   vpip;   /// edge with weight
typedef long long               ll;

#define mp  make_pair
#define ff  first
#define ss  second
#define pb  push_back
#define sz(x)   (int)(x).size()

const int MOD = 1e9+7;  // 998244353;
const int MX  = 2e5+5;   //
const ll  INF = 1e18;    //

#define MAXV 5007
#define MAXE 100007


bool debug;

int N, M;
vii adjlist[MAXV];
int useNode[MAXV];
int dist[MAXV], cnt[MAXV], inq[MAXV];

bool spfa(int s) {  /// Shortest Path Faster Algorithm
    for(int i=0; i<N; i++) {
        dist[i] = MX;
        cnt[i] = 0; inq[i] = 0;
    }
    queue<int> q;

    dist[s] = 0;
    q.push(s); inq[s] = 1;
    while (!q.empty()) {
        int v = q.front();
        q.pop(); inq[v] = 0;

        for(auto e : adjlist[v]) {
            int u = e.ff, w = e.ss;
            if(dist[v] + w < dist[u]) {
                dist[u] = dist[v] + w;
                if(dist[u] < 0 ) return false; /// optimization for TLE.
                if(!inq[u]) {
                    q.push(u); inq[u] = 1;
                    cnt[u]++;
                    if(cnt[u]>N) return false;
                }
            }
        }
    }

    return true;
}

int main() {
    debug = false;
    ios_base::sync_with_stdio(false); cin.tie(0);

    cin >> N >> M;

    /// solve linear inequality function using negative weight SSSP approach
    for(int i=0;i<M; i++) {
        int l, r, k, val; cin >> l >> r >> k >> val;
        /// xi is number of 1 from 0 to i
        l++; r++;
        if(val == 1) {
            /// k-th smallest is 1, i.e. # of 0 < k
            /// (r-l+1) - (x[r] - x[l-1]) < k, i.e. x[l-1] - x[r] <= -(r-l+1) + k -1
            adjlist[r].pb(mp(l-1, -(r-l+1) + k -1));
            if(debug) cout << r << "->" << l-1 << " = " << -(r-l+1) + k -1 << endl;
        } else {
            /// k-th smallest is 0, i.e. # of 0 >= k
            /// (r-l+1) - (x[r] - x[l-1]) >= k, i.e. x[r] - x[l-1] <= (r-l+1)-k
            adjlist[l-1].pb(mp(r, (r-l+1)-k));
            if(debug) cout << l-1 << "->" << r << " = " << k -1 << endl;
        }
    }
    /// restriction: the increase from x[i-1] to x[i] is either 0 or 1
    for(int i=1; i<=N; i++) {
        /// x[i] - x[i-1] <=1
        adjlist[i-1].pb(mp(i, 1));
        /// x[i] - x[i-1] >=0, i.e. x[i-1] - x[i] <=0
        adjlist[i].pb(mp(i-1, 0));
    }
    N++;

    /// SPFA
    if(!spfa(0)) {
        cout << -1 << endl;
        if(debug) {
            cout << endl;
            for(int i=1; i<N; i++) cout << dist[i] << endl;
        }
        exit(0);
    }

    /// output a possible solution which is the distance from source s
    for(int i=1; i<N; i++) {
        if(dist[i] > dist[i-1]) cout << 1;
        else cout << 0;
        cout << " ";
    }
    cout << endl;

    if(debug) cout << endl << "EOL" << endl;

}

• Subtask 1: 7 / 7
• Subtask 2: 13 / 13
• Subtask 3: 25 / 25
• Subtask 4: 55 / 55