Sparse Graph

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

Total Submission(s): 2612    Accepted Submission(s): 911

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 not adjacent 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N个顶点，M条边的无向图G，求其补图H中给定的源点S到其他N-1个点的最短距离，每条边的长度为单位长度1。

 

解法：利用补图求最短路，由于点比较多，所以利用链式前向星存图，建图后将已存在边标记，然后BFS收缩补图。

 

补图：图G的补图，通俗的来讲就是完全图K

_n去除G的边集后得到的图K

_n-G。在 里面，一个图

G的

补图（complement）或者 （inverse）是一个图有着跟

G相同的点，而且这些点之间有边相连 在

G里面他们没有边相连。

的补图就是

 

对于BFS我们这样编写

void bfs(int start,int n){    set
      
        s1;    set
       
         s2;    set
        
         ::iterator it;        for(int i=1;i<=n;i++)        {            if(i!=start)            s1.insert(i);        }    dis[start]=0;    queue
         
           que;    que.push(start);    while(!que.empty())    {        int u=que.front();        que.pop();        for(int i=head[u];i!=-1;i=edge[i].next)        {            int v=edge[i].in;            if(!s1.count(v))continue;            s1.erase(v);            s2.insert(v);        }        for(it=s1.begin();it!=s1.end();it++)        {            que.push(*it);            dis[*it]=dis[u]+1;        }        s1.swap(s2);        s2.clear();    }}
         
        
       
      

利用两个set容器进行对已求得最短路的点、未求得的点和连接了原图边的点的区分和筛选。

例如对上诉图

先将3和2编入再依次对剩下的点进行搜索。

 

 

 最后得到定点搭配其他点的最短路

全部代码如下

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