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 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

Source

2016 ACM/ICPC Asia Regional Dalian Online

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
const int N=1e5+10,M=4e6+10,inf=1e9+10,mod=1e9+7;
const ll INF=1e18+10;
vector<int>v[N<<1];
queue<int>q;
set<int>s;
int flag[N<<1];
int ans[N<<1];
void init(int n)
{
    s.clear();
    for(int i=1;i<=n;i++)
        v[i].clear(),flag[i]=1,ans[i]=0;
    while(!q.empty())q.pop();
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        init(n);
        for(int i=0;i<m;i++)
        {
            int u,w;
            scanf("%d%d",&u,&w);
            v[u].push_back(w);
            v[w].push_back(u);
        }
        int st;
        scanf("%d",&st);
        q.push(st);
        for(int i=1;i<=n;i++)if(i!=st)s.insert(i);
        flag[st]=0,ans[st]=0;
        while(!q.empty())
        {
            int u=q.front();
            q.pop();
            for(int i=0;i<v[u].size();i++)
                flag[v[u][i]]=0;
            for(set<int>::iterator itt,it=s.begin();it!=s.end();)
            {
                if(flag[*it])
                {
                    q.push(*it);
                    ans[*it]=ans[u]+1;
                    itt=it;
                    it++;
                    s.erase(itt);
                }
                else
                    it++;
            }
            for(set<int>::iterator it=s.begin();it!=s.end();it++)flag[*it]=1;
        }
        int flag=0;
        for(int i=1;i<=n;i++)
        {
            if(i==st)
            continue;
            printf("%d%c",ans[i],(flag++>=(n-2)?‘\n‘:‘ ‘));
        }
    }
    return 0;
}