hdu Sparse Graph(补集）

Sparse Graph

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2487    Accepted Submission(s): 386

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 inG . 

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\le T<35) denoting the number of test cases. For each test case, the first line contains two integers N(2\le N\le 200000) and M(0\le M\le 20000) . The following M lines each contains two distinct integers u, v(1\le u,v\le N) denoting an edge. And S\ (1\le S\le 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

12 01

 

Sample Output

1
这题是要求补集，创新度很大，，，容易超时，，，
#include<cstdio>#include<iostream>#include<cstring>#include<queue>#include<algorithm>#include<cmath>#include<map>#include<set>#include<vector>using namespace std;const int N = 4000010;int dis[N], visit[N], head[N];vector<int>G[N];const int inf = 2e6+7;typedef long long LL;void add(int u, int v);int cnt;struct node{    int next, to;}p1[N];int main(){    int t;    scanf("%d", &t);    while(t--)    {        int n, m;        scanf("%d %d", &n, &m);        memset(head,-1,sizeof(head));        cnt=0;        for(int i=0;i<m;i++)        {            int u, v;            scanf("%d %d", &u, &v);            add(u,v);            add(v,u);        }        int s;        scanf("%d", &s);        for(int i=1;i<=n;i++)        {            dis[i]=inf;        }        dis[s]=0;        set<int>q1, q2;        set<int>::iterator it;        q1.clear(), q2.clear();        for(int i=1;i<=n;i++)        {            if(i!=s)            {                q1.insert(i);            }        }        queue<int>q;        while(!q.empty())        {            q.pop();        }        q.push(s);        while(!q.empty())        {            int u=q.front();            q.pop();            for(int i=head[u];i!=-1;i=p1[i].next)            {                int v=p1[i].to;                if(!q1.count(v))                    continue;                q1.erase(v);                q2.insert(v);            }            for(it=q1.begin();it!=q1.end();it++)            {                dis[*it]=dis[u]+1;                q.push(*it);            }            q1.swap(q2),q2.clear();        }        for(int i=1;i<=n;i++)        {            if(i!=s)            {                printf("%d%c",dis[i]==inf?-1:dis[i]," \n"[i==n]);            }        }    }    return 0;}void add(int u, int v){    p1[cnt].to=v;    p1[cnt].next=head[u];    head[u]=cnt++;    return ;}