CodeForces 1196F K-th Path

感想

集训了两个星期，感觉自己水平也在不断提高，挺好。昨天补了五道CF题，今天来更新博客。

这题是CodeForces1196，一场div3的最后一题。

中文题意

给一张\(n\)个点\(m\)条边的无向连通图，让求这张图上所有最短路中，第\(k\)短的最短路（交换起点终点算同一条路）。

解题思路

官方题解是这么说的——

1196F - K-th Path

The main observation is that you don't need more than min(k,m) $min(k,m)$ smallest by weight edges (among all edges with the maximum weights you can choose any). Maybe there will be a proof later, but now I ask other participant to write it.

So you sort the initial edges and after that you can construct a graph consisting of no more than 2min(k,m) $2min(k,m)$ vertices and no more than min(m,k) $min(m,k)$ edges. You just can build the new graph consisting only on these vertices and edges and run Floyd-Warshall algorithm to find the matrix of shortest paths. Then sort all shorted distances and print the k $k$ -th element of this sorted array.

Time complexity: O(mlogm+k3) $O(m\log m+k^3)$ .

I know that there are other approaches that can solve this problem with greater k $k$ , but to make this problem easily this solution is enough.

哈哈