time. Find the cycle created by the edge, and contract the entire cycle into one vertex. Repeat for the entire Dijkstra run. This should take the same total time complexity with Dijkstra. The edges in the shortest path of the final graph are the ones in the intersection of all shortest ...
else { parent[y] = x; // 父节点 x // 并查集检查环 if (!uf.union(x, y)) cycle = i; } } if (conflict < 0) return edges[cycle]; else { int[] conflictEdge = edges[conflict]; if (cycle >= 0) conflictEdge[0] = parent[conflictEdge[1]]; return conflictEdge; } } } 1. ...
0133-clone-graph.py 0134-gas-station.py 0136-single-number.py 0138-copy-list-with-random-pointer.py 0139-word-break.py 0141-linked-list-cycle.py 0143-reorder-list.py 0146-lru-cache.py 0150-evaluate-reverse-polish-notation.py 0152-maximum-product-subarray.py 0153-fin...