hdu-5876-Sparse Graph

Sparse Graph

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)


Problem Description
In graph theory, the complement of a graph G is a graph H on the same vertices such that two distinct vertices of H are adjacent if and only if they are notadjacent in G. 


Now you are given an undirected graph G of N nodes and M bidirectional edges of unit length. Consider the complement of G, i.e., H. For a given vertex S on H, you are required to compute the shortest distances from S to all N−1 other vertices.
 


Input
There are multiple test cases. The first line of input is an integer T(1≤T<35) denoting the number of test cases. For each test case, the first line contains two integers N(2≤N≤200000) and M(0≤M≤20000). The following M lines each contains two distinct integers u,v(1≤u,v≤N) denoting an edge. And S (1≤S≤N) is given on the last line.
 


Output
For each of T test cases, print a single line consisting of N−1 space separated integers, denoting shortest distances of the remaining N−1 vertices from S (if a vertex cannot be reached from S, output ``-1" (without quotes) instead) in ascending order of vertex number.
 

Sample Input
1
2 0
1
 
Sample Output
1

题意：


给定一个具有n个点，m条边的无向图G，求在其“补图”中一起点st到其余所有定点的最短距离。
题目链接：Sparse Graph


解题思路：
注意到G图中最多只有200 00条边，但定点可达2 000 00,是稀疏图，因此还原其补图H的话肯定受不了时间，
考虑枚举G图中的边，只要任意点u到v没边，那么在H中就是有边的，因此只要处理G图中有边的顶点，然后从已
经处理好的顶点Dijkstra求最短路即可，


但G图中的边少，意味着H图中很多点之间都是有边的，那么循环过程中可以很快就能求除答案，需要提前退出循
环，cnt记录还需处理的顶点个数


代码：

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