https://codeforces.com/contest/545/problem/E [算法] 首先求 u 到所有结点的最短路 记录每个节点最短路径上的最后一条边 答案即为以u为根的一棵最短路径生成树 时间复杂度 : O(NlogN) [代码] #include<bits/stdc++.h>usingnamespacestd;constintMAXN = 3e5 +10;constlonglongINF =1e60;structedge ...
In the first sample there are two possible shortest path trees: with edges 1 – 3 and 2 – 3 (the total weight is 3); with edges 1 – 2 and 2 – 3 (the total weight is 2); And, for example, a tree with edges 1 – 2 and 1 – 3 won't be a...
https://codeforces.com/contest/545/problem/E [算法] 首先求 u 到所有结点的最短路 记录每个节点最短路径上的最后一条边 答案即为以u为根的一棵最短路径生成树 时间复杂度 : O(NlogN) [代码] #include<bits/stdc++.h>usingnamespacestd;constintMAXN = 3e5 +10;constlonglongINF =1e60;structedge ...
E) (hereVis the set of vertices,Eis the set of edges). The shortest-path tree from vertexuis such graphG1 = (V, E1) that is a tree with the set of edgesE1 that is the subset of the set of edges of the initial graphE, and the lengths of the shortest paths fromuto ...
A tree decomposition in paths is valid if each vertex belongs to exactly one path. A path can be a single vertex. Two decompositions are different if there are two vertices and that belong to the same path in one of the decompositions but two distinct paths in the other. ...
For both ways you divide the problem into two parts. The first path is to erase the cycle and for every tree in the resulting forest to find the diameter. Now in the second part we should only consider vertices from different trees in the forest. Obviously for every tree it will be opt...
In the first sample there are two possible shortest path trees: with edges1 – 3 and2 – 3 (the total weight is3); with edges1 – 2 and2 – 3 (the total weight is2); And, for example, a tree with edges1 – 2 and1 – 3 won't be a shortest path...
d, c, n, e=hq.heappop(q)ifvis[n]:continueans.append(e) tot+=c vis[n]=Trueforv, c, iinadj[n]:ifnotvis[v]: hq.heappush(q, (d+c, c, v, i)) ans=map(str, ans)printtotprint"".join(ans[1:]) 上面的代码都是在codeforces上面抄过来的,自己写不出来。。。