이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
*/
using namespace std;
// using namespace __gnu_pbds;
using namespace chrono;
// mt19937 rng((int) std::chrono::steady_clock::now().time_since_epoch().count());
/*
template <class T> using ordered_set = tree <T, null_type, less <T>, rb_tree_tag, tree_order_statistics_node_update>;
*/
//***************** CONSTANTS *****************
const int MXN = 50001;
const int LOG = 16;
//***************** GLOBAL VARIABLES *****************
int dp[MXN][LOG][5][5];
//***************** AUXILIARY STRUCTS *****************
//***************** MAIN BODY *****************
void solve(){
int K, N, M, O;
cin >> K >> N >> M >> O;
int B = N / K;
memset(dp, 0x3f, sizeof dp);
for(int i = 0; i < M; i++){
int u, v, t;
cin >> u >> v >> t;
dp[u / K][0][u % K][v % K] = t;
}
for(int j = 1; j < LOG; j++){
for(int i = 0; i <= B; i++){
for(int f = 0; f < K; f++)
for(int t = 0; t < K; t++){
if(i + (1 << j) > B) continue;
for(int m = 0; m < K; m++)
dp[i][j][f][t] = min(dp[i][j][f][t], dp[i][j-1][f][m] + dp[i + (1 << (j - 1))][j-1][m][t]);
}
}
}
for(int _ = 0; _ < O; _++){
int u, v;
cin >> u >> v;
int x = u / K, y = v / K;
int res[K];
memset(res, 0x3f, sizeof res);
res[u % K] = 0;
for(int j = LOG - 1; j >= 0; --j){
if(x + (1 << j) <= y){
int cur[K];
memset(cur, 0x3f, sizeof cur);
for(int t = 0; t < K; t++)
for(int f = 0; f < K; f++)
cur[t] = min(cur[t], res[f] + dp[x][j][f][t]);
for(int i = 0; i < K; i++)
res[i] = cur[i];
x += (1 << j);
}
}
cout << (res[v % K] == 1061109567 ? -1 : res[v % K]) << '\n';
}
}
//***************** *****************
int32_t main(){
ios_base::sync_with_stdio(NULL);
cin.tie(NULL);
#ifdef LOCAL
auto begin = high_resolution_clock::now();
#endif
int tc = 1;
// cin >> tc;
for (int t = 0; t < tc; t++)
solve();
#ifdef LOCAL
auto end = high_resolution_clock::now();
cout << fixed << setprecision(4);
cout << "Execution Time: " << duration_cast<duration<double>>(end - begin).count() << "seconds" << endl;
#endif
return 0;
}
/*
If code gives a WA, check for the following :
1. I/O format
2. Are you clearing all global variables in between tests if multitests are a thing
3. Can you definitively prove the logic
4. If the code gives an inexplicable TLE, and you are sure you have the best possible complexity,
use faster io
*/
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