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>
using namespace std;
typedef long long ll;
typedef long double ld;
#define rep(a, b) for(int a = 0; a < (b); ++a)
#define st first
#define nd second
#define pb push_back
#define all(a) a.begin(), a.end()
const int LIM=1e5+7;
const ll INF=1e18+7;
vector<pair<int,ll>>V[LIM];
ll wart[LIM][2], odl[LIM][2], mi[LIM][2], n, m;
void dijkstra1(int x, int k) {
rep(i, n) wart[i][k]=INF;
priority_queue<pair<ll,int>>q;
q.push({0, x});
while(!q.empty()) {
ll o=-q.top().st, p=q.top().nd; q.pop();
if(wart[p][k]<=o) continue;
wart[p][k]=o;
for(auto i : V[p]) if(o+i.nd<wart[i.st][k]) {
q.push({-o-i.nd, i.st});
}
}
}
void dijkstra2(int x, int k) {
rep(i, n) odl[i][k]=mi[i][k]=INF;
priority_queue<pair<pair<ll,ll>,int>>q;
q.push({{0, -wart[x][0]}, x});
while(!q.empty()) {
ll a=-q.top().st.st, b=-q.top().st.nd, p=q.top().nd; q.pop();
if(odl[p][k]<a) continue;
if(odl[p][k]==a && mi[p][k]<=b) continue;
odl[p][k]=a;
mi[p][k]=b;
for(auto i : V[p]) {
if(a+i.nd<odl[i.st][k] || a+i.nd==odl[i.st][k] && min(b, wart[i.st][0])<mi[i.st][0]) {
q.push({{-a-i.nd, -min(b, wart[i.st][0])}, i.st});
}
}
}
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
int s, t, u, v;
cin >> n >> m >> s >> t >> u >> v;
--s; --t; --u; --v;
rep(i, m) {
int a, b, c;
cin >> a >> b >> c; --a; --b;
V[a].pb({b, c});
V[b].pb({a, c});
}
dijkstra1(u, 0);
dijkstra1(v, 1);
dijkstra2(s, 0);
dijkstra2(t, 1);
ll ans=wart[v][0];
rep(i, n) if(odl[i][0]+odl[i][1]==odl[t][0]) {
ans=min(ans, wart[i][1]+min(mi[i][0], mi[i][1]));
}
cout << ans << '\n';
}
컴파일 시 표준 에러 (stderr) 메시지
commuter_pass.cpp: In function 'void dijkstra2(int, int)':
commuter_pass.cpp:38:51: warning: suggest parentheses around '&&' within '||' [-Wparentheses]
38 | if(a+i.nd<odl[i.st][k] || a+i.nd==odl[i.st][k] && min(b, wart[i.st][0])<mi[i.st][0]) {
| ~~~~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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