HDU 5876 Sparse Graph(补图+BFS最短路）

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Sparse Graph

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 617 Accepted Submission(s): 210

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：给你一个图，和一个点s,然后问在它的补图中，点s到每个点的最短路是多长。

解题思路： 题意不难，但是第一次在树上做BFS，感觉边的关系不好维护，比赛时想到一个想法，但是不敢想，赛后写出来了。
给每个点开一个vector，然后记录跟这个点直接连接的点。然后以s为起点进行bfs.

<span style="font-size:18px;"> #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<set>
 #include<queue>
 using namespace std;
 const int maxn=200010;
 vector<int> v[maxn];
 int n,m,dis[maxn],s;
 void BFS()
 {
     set<int> s1,s2;
     set<int>::iterator it;
     queue<int> que;
     que.push(s);
     dis[s]=0;
     for(int i=1;i<=n;i++)
        if(i!=s) s1.insert(i);


     while(!que.empty())
     {
         int u=que.front();
         que.pop();
         int len=v[u].size();
         for(int i=0;i<len;i++)
         {
             int vv=v[u][i];
             if(!s1.count(vv))
                continue;
             s1.erase(vv);
             s2.insert(vv);
         }
         for(it=s1.begin();it!=s1.end();it++)
         {
             que.push(*it);
             dis[*it]=dis[u]+1;
         }
         s1.swap(s2);
         s2.clear();
     }
 }
 int main()
 {
     int t,a,b;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d %d",&n,&m);
         memset(dis,0x3f,sizeof(dis));
         for(int i=0;i<maxn;i++) v[i].clear();
         for(int i=0;i<m;i++)
         {
             scanf("%d %d",&a,&b);
             v[a].push_back(b);
             v[b].push_back(a);
         }
         scanf("%d",&s);
         BFS();
         for(int i=1;i<=n;i++)
         {
             if(i==s){
                if(i==n) printf("\n");
                continue;
             }
             printf("%d%c",dis[i],i==n? '\n' : ' ');
         }
     }
     return 0;
 }
</span>

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