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 "cyberland.h"
#include <vector>
using namespace std;
using pid = pair<int, double>;
using tp = tuple<double, int, int, int>;
const int N = 1e5 + 5, K = 70;
double d[N][K][2], ans = DBL_MAX;
vector<pid> v[N];
bool f[N][K][2];
priority_queue<tp, vector<tp>, greater<tp>> q;
double solve(int n, int m, int k, int h, std::vector<int> x, std::vector<int> y, std::vector<int> c, std::vector<int> arr) {
k = min(k, 67); ans = DBL_MAX;
for (int i = 0; i <= n; i++) for (int j = 0; j <= k; j++) d[i][j][0] = d[i][j][1] = 1e11, f[i][j][0] = f[i][j][1] = 0;
for (int i = 0; i <= n; i++) v[i].clear();
while (!q.empty()) q.pop();
for (int i = 0; i < m; i++) v[x[i]].emplace_back(y[i], c[i]), v[y[i]].emplace_back(x[i], c[i]);
q.emplace(d[h][0][0] = 0, h, 0, 0);
while (!q.empty()) {
auto [nd, nn, nk, ns] = q.top(); q.pop();
if (f[nn][nk][ns]) continue; f[nn][nk][ns] = 1;
for (auto &[tn, td]: v[nn]) {
double ntd = d[nn][nk][ns] + (!ns) * (td / pow(2.00, nk));
if (d[tn][nk][ns] > ntd) q.emplace(d[tn][nk][ns] = ntd, tn, nk, ns);
}
if (!arr[nn] && !ns) for (auto &[tn, td]: v[nn]) if (d[tn][nk][1] > nd) q.emplace(d[tn][nk][1] = nd, tn, nk, 1);
if (arr[nn] == 2 && nk < k) {
for (auto &[tn, td]: v[nn]) {
double ntd = d[nn][nk][ns] + (!ns) * (td / pow(2.00, nk + 1));
if (d[tn][nk + 1][ns] > ntd) q.emplace(d[tn][nk + 1][ns] = ntd, tn, nk + 1, ns);
}
}
}
for (int i = 0; i <= k; i++) for (int j = 0; j < 2; j++) ans = min(ans, d[0][i][j]);
return (ans == DBL_MAX ? -1 : ans);
}
컴파일 시 표준 에러 (stderr) 메시지
cyberland.cpp: In function 'double solve(int, int, int, int, std::vector<int>, std::vector<int>, std::vector<int>, std::vector<int>)':
cyberland.cpp:24:9: warning: this 'if' clause does not guard... [-Wmisleading-indentation]
24 | if (f[nn][nk][ns]) continue; f[nn][nk][ns] = 1;
| ^~
cyberland.cpp:24:38: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'if'
24 | if (f[nn][nk][ns]) continue; f[nn][nk][ns] = 1;
| ^
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