Submission #28043

#	Time	Username	Problem	Language	Result	Execution time	Memory
28043		sampriti	Jakarta Skyscrapers (APIO15_skyscraper)	C++14	57 / 100	869 ms	261272 KiB

skyscraper

#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <climits>

using namespace std;

#define LIM 150

int B[30000], P[30000];
set<int> A[30000];
vector<pair<pair<short, short>, unsigned char> > G[30000][LIM + 1];
short D[30000][LIM + 1];
set<pair<short, pair<short, short> > > Q;

int main() {
  ios::sync_with_stdio(false); cin.tie(0);

  int N, M; cin >> N >> M;

  for(int i = 0; i < M; i++) {
    cin >> B[i] >> P[i];
    A[B[i]].insert(P[i]);
  }

  for(int i = 0; i < N; i++) {
    for(int x: A[i]) {
      if(x < LIM) {
        G[i][LIM].push_back({{i, x}, 0});
      }
    }

    for(int x: A[i]) {
      if(x >= LIM) {
        for(int d = 1; i + d * x < N; d++) {
          if(A[i + d * x].size() == 0) continue;
          G[i][LIM].push_back({{i + d * x, LIM}, d});
        }
        for(int d = 1; i - d * x >= 0; d++) {
          if(A[i - d * x].size() == 0) continue;
          G[i][LIM].push_back({{i - d * x, LIM}, d});
        }
      }
    }

    for(int x = 1; x < LIM; x++) {
      G[i][x].push_back({{i, LIM}, 0});

      if(i + x < N) {
        G[i][x].push_back({{i + x, x}, 1});
      }
      if(i - x >= 0) {
        G[i][x].push_back({{i - x, x}, 1});
      }
    }
  }

  for(int i = 0; i < N; i++) {
    for(int j = 0; j <= LIM; j++) {
      D[i][j] = SHRT_MAX;
    }
  }

  Q.insert({0, {B[0], LIM}}); D[B[0]][LIM] = 0;

  while(!Q.empty()) {
    auto it = *Q.begin(); Q.erase(Q.begin());
    int i = it.second.first, j = it.second.second;

    if(i == B[1] && j == LIM) break;

    for(auto it2: G[i][j]) {
      int i2 = it2.first.first, j2 = it2.first.second, d = it2.second;
      if(D[i][j] + d < D[i2][j2]) {
        if(D[i2][j2] != SHRT_MAX) {
          Q.erase({D[i2][j2], {i2, j2}});
        }
        D[i2][j2] = D[i][j] + d;
        Q.insert({D[i2][j2], {i2, j2}});
      }
    }
  }

  if(D[B[1]][LIM] == SHRT_MAX) {
    cout << -1 << endl;
  }
  else {
    cout << D[B[1]][LIM] << endl;
  }
}

• Subtask 1: 10 / 10
• Subtask 2: 12 / 12
• Subtask 3: 14 / 14
• Subtask 4: 21 / 21
• Subtask 5: 0 / 43