This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
#define int long long
#define rep(i,j,k) for (int i=j; i<=k; i++)
#define de(x) cout << #x << "=" << x << ", "
#define dd cout << '\n';
#define ff first
#define ss second
#define pb push_back
#define lyx_my_wife ios::sync_with_stdio(0), cin.tie(0);
using namespace std;
typedef pair<int,int> pii;
const int N = 1e5+10, M = 2e5+10;
int n, m, s, t, u, v;
vector<pii> G[N];
int disu[N], disv[N], diss[N], dist[N], vis[N];
int dpu[N], dpv[N], ans;
void dijks(int rt, int dis[N], int fordp=0, int forans=0)
{
memset(dis, 0x3f, sizeof(disu));
memset(vis, 0, sizeof(vis));
priority_queue<pii, vector<pii>, greater<pii>> pq;
dis[rt] = 0;
pq.push({0, rt});
if (fordp)
{
memset(dpu, 0x3f, sizeof(dpu));
memset(dpv, 0x3f, sizeof(dpv));
}
while(pq.size())
{
int i = pq.top().ss, d = pq.top().ff;
pq.pop();
if (vis[i]) continue;
vis[i] = 1;
if (fordp)
{
dpu[i] = min(dpu[i], disu[i]);
dpv[i] = min(dpv[i], disv[i]);
}
for (auto e: G[i])
{
int j = e.ss, w = e.ff;
if (forans)
{
if (diss[j] != diss[i]-w) continue;
ans = min(ans, dpu[j]+disv[i]);
ans = min(ans, dpv[j]+disu[i]);
}
if (dis[j] > d+w)
{
dis[j] = d+w;
pq.push({dis[j], j});
if (fordp) dpu[j] = dpu[i], dpv[j] = dpv[i];
}
else if (dis[j] == d+w)
{
if (fordp) dpu[j] = min(dpu[j], dpu[i]), dpv[j] = min(dpv[j], dpv[i]);
}
}
}
}
signed main()
{
lyx_my_wife
cin >> n >> m >> s >> t >> u >> v;
rep(_,1,m)
{
int i, j, w;
cin >> i >> j >> w;
G[i].pb({w,j});
G[j].pb({w,i});
}
// dijks for disu, disv
dijks(u,disu);
dijks(v,disv);
dijks(s,diss,1);
ans = disu[v];
dijks(t,dist,0,1);
cout << ans << '\n';
}
/*
6 6
1 6
1 4
1 2 1
2 3 1
3 5 1
2 4 3
4 5 2
5 6 1
6 5
1 2
3 6
1 2 10
2 3 10
3 4 10
4 5 10
5 6 10
8 8
5 7
6 8
1 2 2
2 3 3
3 4 4
1 4 1
1 5 5
2 6 6
3 7 7
4 8 8
5 5
1 5
2 3
1 2 1
2 3 10
2 4 10
3 5 10
4 5 10
*/
/*
> 無向圖,選一條 s-t 最短路使其邊權歸零,使 u-v 路徑最短
將無向圖拆成有向圖,從 s 開始的最短路徑有一個 DAG as sDAG
s->t最短路也有 DAG 屬於 sDAG as stDAG
! stDAG 的任何一條 st 路徑都是合法的最短路
! stDAG inverse as tsDAG 的任何一條 ts 路徑都是合法的最短路
from u to v in stDAG
from v to u in stDAG
! u -> (x -> y)stDAG -> v
! sv -> (x -> y)stDAG -> u
! ans = min(dis[u][x] + dis[y][v])
! ans = min(dis[v][x] + dis[y][u])
! ans = min(dis[u][x] + dis[v][y])
! ans = min(dis[v][x] + dis[u][y])
+ dpu: on stDAG, min dis[u][x]
+ dpv: on stDAG, min dis[v][x]
+ ans = min(ans, dpu + disv[i])
+ ans = min(ans, dpv + disu[i])
*/
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