2203. Minimum Weighted Subgraph With the Required Paths
You are given an integer n
denoting the number of nodes of a weighted directed graph. The nodes are numbered from 0
to n - 1
.
You are also given a 2D integer array edges
where edges[i] = [fromi, toi, weighti]
denotes that there exists a directed edge from fromi
to toi
with weight weighti
.
Lastly, you are given three distinct integers src1
, src2
, and dest
denoting three distinct nodes of the graph.
Return the minimum weight of a subgraph of the graph such that it is possible to reach dest
from both src1
and src2
via a set of edges of this subgraph. In case such a subgraph does not exist, return -1
.
A subgraph is a graph whose vertices and edges are subsets of the original graph. The weight of a subgraph is the sum of weights of its constituent edges.
Example 1:

Input: n = 6, edges = [[0,2,2],[0,5,6],[1,0,3],[1,4,5],[2,1,1],[2,3,3],[2,3,4],[3,4,2],[4,5,1]], src1 = 0, src2 = 1, dest = 5 Output: 9 Explanation: The above figure represents the input graph. The blue edges represent one of the subgraphs that yield the optimal answer. Note that the subgraph [[1,0,3],[0,5,6]] also yields the optimal answer. It is not possible to get a subgraph with less weight satisfying all the constraints.
Example 2:

Input: n = 3, edges = [[0,1,1],[2,1,1]], src1 = 0, src2 = 1, dest = 2 Output: -1 Explanation: The above figure represents the input graph. It can be seen that there does not exist any path from node 1 to node 2, hence there are no subgraphs satisfying all the constraints.
Constraints:
3 <= n <= 105
0 <= edges.length <= 105
edges[i].length == 3
0 <= fromi, toi, src1, src2, dest <= n - 1
fromi != toi
src1
,src2
, anddest
are pairwise distinct.1 <= weight[i] <= 105
class Solution { | |
public: | |
struct Edge { | |
int to; | |
int cost; | |
}; | |
void dijkstra(vector<vector<Edge>> &G, int s, vector<long long>& dis){ | |
// get min from queue | |
priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> pq; | |
dis[s] = 0; | |
pq.emplace(dis[s], s); | |
while (!pq.empty()) { | |
auto p = pq.top(); | |
pq.pop(); | |
// next | |
int v = p.second; | |
if (dis[v] < p.first) { | |
continue; | |
} | |
// 最短距離を更新 | |
for (auto &e : G[v]) { | |
// 最短距離候補なら priority_queue に追加 | |
if (dis[e.to] > dis[v] + e.cost) { | |
dis[e.to] = dis[v] + e.cost; | |
pq.emplace(dis[e.to], e.to); | |
} | |
} | |
} | |
} | |
long long minimumWeight(int n, vector<vector<int>>& edges, int src1, int src2, int dest) { | |
// make graph | |
vector<vector<Edge>> G(n), R(n); | |
for(auto e: edges){ | |
// from -> to | |
G[e[0]].push_back(Edge({e[1], e[2]})); | |
// to -> from | |
R[e[1]].push_back(Edge({e[0], e[2]})); | |
} | |
// sr2 -> others, src2 -> others | |
vector<long long> dis1(n, LONG_MAX), dis2(n, LONG_MAX), disR(n, LONG_MAX); | |
dijkstra(G, src1, dis1); | |
dijkstra(G, src2, dis2); | |
// dest -> others | |
dijkstra(R, dest, disR); | |
// min ans | |
long long ans = LONG_MAX; | |
for(int i = 0; i < n; i++){ | |
if(dis1[i] != LONG_MAX and dis2[i] != LONG_MAX and disR[i] != LONG_MAX) | |
ans = min(ans, dis1[i] + dis2[i] + disR[i]); | |
} | |
if(ans == LONG_MAX) | |
return -1; | |
return ans; | |
} | |
}; |
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