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;
#define faster ios_base::sync_with_stdio(0); cin.tie(0);
#define FOR(i,a,b) for(int i = a; i <= b; i++)
#define FOD(i,a,b) for(int i = a; i >= b; i--)
#define int long long
#define fi first
#define se second
#define pb push_back
#define ll long long
#define ull unsigned long long
#define lcm(a,b) (a*b)/__gcd(a,b)
#define ii pair<int,int>
#define iii pair<int,pair<int,int>>
#define iv pair<pair<int,int>,pair<int,int>>
const int inf = 1e9;
const ll INF = 1e18;
const int mod = 1e9 + 7;
const int N = 1e5 + 105;
const int dx[] = {-1,0,1,0};
const int dy[] = {0,1,0,-1};
const int dxx[] = {-1,-1,0,1,1,1,0,-1};
const int dyy[] = {0,1,1,1,0,-1,-1,-1};
int n, m, S, T, U, V, ans;
vector<ii> g[N];
int du[N], dv[N], dp[N][2], ds[N];
void dijk1(int beg, int d[])
{
priority_queue<ii,vector<ii>,greater<ii>> pq;
FOR(i,1,n) d[i] = INF;
d[beg] = 0;
pq.push({0,beg});
while(!pq.empty())
{
int u = pq.top().se;
int cost = pq.top().fi;
pq.pop();
if(cost > d[u])
continue;
for(ii H : g[u])
{
int v = H.fi;
int l = H.se;
if(d[v] > d[u] + l)
{
d[v] = d[u] + l;
pq.push({d[v],v});
}
}
}
}
void dijk2(int beg, int des)
{
priority_queue<iii,vector<iii>,greater<iii>> pq;
pq.push({0,{0,beg}});
FOR(i,0,n)
{
dp[i][0] = INF;
dp[i][1] = INF;
ds[i] = INF;
}
while(!pq.empty())
{
int u = pq.top().se.se;
int par = pq.top().se.fi;
int cost = pq.top().fi;
pq.pop();
if(ds[u] == INF)
{
ds[u] = cost;
dp[u][0] = min(dp[par][0], du[u]);
dp[u][1] = min(dp[par][1], dv[u]);
for(ii H : g[u])
pq.push({cost + H.se, {u, H.fi}});
}
else if(ds[u] == cost)
{
if(min(dp[par][0], du[u]) + min(dp[par][1], dv[u]) <= dp[u][0] + dp[u][1])
{
dp[u][0] = min(dp[par][0], du[u]);
dp[u][1] = min(dp[par][1], dv[u]);
}
}
}
ans = min(ans, dp[des][0] + dp[des][1]);
}
void solve()
{
dijk1(U,du);
dijk1(V,dv);
ans = du[V];
dijk2(S,T);
dijk2(T,S);
cout << ans;
}
void input()
{
//freopen("TEST.INP", "r", stdin);
//freopen("TEST.OUT", "w", stdout);
cin >> n >> m >> S >> T >> U >> V;
FOR(i,1,m)
{
int u, v, w;
cin >> u >> v >> w;
g[u].pb({v,w});
g[v].pb({u,w});
}
}
signed main()
{
faster
input();
solve();
}
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