HDU 5876 Sparse Graph【补图BFS+set】

Sparse Graph

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1604    Accepted Submission(s): 565

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

12 01

 

Sample Output

1

 

Source

2016 ACM/ICPC Asia Regional Dalian Online

思路：

     维护一个集合ac，在当前集合中表示可以拓展到点。
     维护一个集合nac，在当前集合中表示不可以拓展到点。
 bfs搜索，从当前队列中取出一点u，将图G中与u相连的边从ac中删去，并添加到nac中，并拓展ac中到点，添加到队列中，如此反复，直到队列为空。

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<climits>#include<string>#include<queue>#include<stack>#include<set>#include<map>#include<algorithm>using namespace std;#define rep(i,j,k)for(i=j;i<k;i++)#define per(i,j,k)for(i=j;i>k;i--)#define MS(x,y)memset(x,y,sizeof(x))typedef long long LL;const int INF=0x7ffffff;const int M=2e5+1;vector<int>g[M];int dis[M];int i,j,k,n,m;int s;void solve(){    set<int>ac,nac;    for(i=1;i<=n;i++){        if(i!=s)ac.insert(i);    }    queue<int>q;    q.push(s);    MS(dis,0);    while(!q.empty())    {        int u=q.front();        q.pop();        for(i=0;i<g[u].size();i++){            int &v=g[u][i];            if(!ac.count(v))continue;           ac.erase(v);           nac.insert(v);        }        for(set<int>::iterator it=ac.begin();it!=ac.end();it++){            dis[*it]=dis[u]+1;            q.push(*it);        }        ac.swap(nac);        nac.clear();    }    int flag=0;    for(i=1;i<=n;i++){        if(i==s)continue;        if(flag)printf(" ");        if(!dis[i])printf("-1");        else printf("%d",dis[i]);        flag=1;    }    printf("\n");}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        for(i=1;i<=n;i++)g[i].clear();        for(i=0;i<m;i++){            int a,b;            scanf("%d%d",&a,&b);            g[a].push_back(b);            g[b].push_back(a);        }        scanf("%d",&s);        solve();    }    return 0;}