# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
1258851 | khoile08 | Crocodile's Underground City (IOI11_crocodile) | C++17 | 0 ms | 0 KiB |
#include <bits/stdc++.h>
using namespace std;
#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 db double
#define lcm(a,b) a / __gcd(a, b) * b
#define ii pair<int,int>
#define iii pair<int,pair<int,int>>
#define iv pair<pair<int,int>,pair<int,int>>
#define sq(a) (a) * (a)
#define MASK(i) (1LL << i)
#define task "task"
const int inf = 1e9;
const ll INF = 1e18;
const int mod = 1e9 + 7;
const int N = 1e5 + 5;
const int M = 1e6 + 5;
int d1[N], d2[N];
vector<ii> g[N];
ll travel_plan(int n, int m, int R[][2], int L[], int k, int P[]) {
FOR(i, 0, m - 1) {
g[R[i][0]].pb({R[i][1], L[i]});
g[R[i][1]].pb({R[i][0], L[i]});
}
priority_queue<ii,vector<ii>,greater<ii>> pq;
FOR(i, 0, n - 1) d1[i] = INF;
FOR(i, 0, k - 1) {
d1[P[i]] = 0;
pq.push({0, P[i]});
}
while(!pq.empty()) {
int cost = pq.top().fi;
int u = pq.top().se;
pq.pop();
if(cost != d1[u]) continue;
for(auto H : g[u]) {
int v = H.fi;
int l = H.se;
if(d1[v] > d1[u] + l) {
d1[v] = d1[u] + l;
pq.push({d1[v], v});
}
}
}
FOR(i, 1, n) d2[i] = INF;
d2[0] = 0;
pq.push({0, 0});
while(!pq.empty()) {
int cost = pq.top().fi;
int u = pq.top().se;
pq.pop();
if(cost != d2[u]) continue;
int best = 0, mn = INF;
for(auto H : g[u]) {
int v = H.fi;
int l = H.se;
if(mn > d2[u] + l + d1[v]) {
mn = d2[u] + l + d1[v];
best = v;
}
}
for(auto H : g[u]) {
int v = H.fi;
int l = H.se;
if(v == best) continue;
if(d2[v] > d2[u] + l) {
d2[v] = d2[u] + l;
pq.push({d2[v], v});
}
}
}
int ans = INF;
FOR(i, 0, k - 1) ans = min(ans, d2[P[i]]);
return ans;
}
signed main() {
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int test = 1;
// cin >> test;
while(test--) {
int n, m, k;
cin >> n >> m >> k;
int R[105][2], L[105], P[105];
FOR(i, 0, m - 1) cin >> R[i][0] >> R[i][1];
FOR(i, 0, m - 1) cin >> L[i];
FOR(i, 0, k - 1) cin >> P[i];
cout << travel_plan(n, m, R, L, k, P);
}
}