HDOJ5876(补图的最短路)

Posted 2020-08-09 vCoders

Sparse Graph

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1563    Accepted Submission(s): 549

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

1

2 0

1

 

 

Sample Output

1

 

思路：边的长度均为1，用bfs。遍历补图中与u相连接的结点v，并将其在全部结点的集合中删除。删除结点用set较快。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
using namespace std;
const int MAXN=200005;
vector<int> arc[MAXN];
set<int> node;
int n,m,s;
int vis[MAXN],d[MAXN];
void bfs(int src)
{
    memset(vis,0,sizeof(vis));
    memset(d,-1,sizeof(d));
    queue<int> que;
    que.push(src);
    vis[src]=1;
    d[src]=0;
    node.erase(src);
    while(!que.empty())
    {
        int u=que.front();que.pop();
        vector<int> vec;
        for(set<int>::iterator it=node.begin();it!=node.end();it++)
        {
            int to=*it;
            if(!vis[to]&&!binary_search(arc[u].begin(),arc[u].end(),to))
            {
                d[to]=d[u]+1;
                vis[to]=1;
                que.push(to);            
                vec.push_back(to);
            }
        }
        for(int i=0;i<vec.size();i++)
        {
            node.erase(vec[i]);
        }
    }    
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        node.clear();
        for(int i=1;i<=n;i++)
        {
            node.insert(i);
            arc[i].clear();
        }
        for(int i=0;i<m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            arc[u].push_back(v);
            arc[v].push_back(u);
        }
        for(int i=1;i<=n;i++)
        {
            sort(arc[i].begin(),arc[i].end());
        }
        scanf("%d",&s);
        bfs(s);
        bool tag=false;
        for(int i=1;i<=n;i++)
        {
            if(i==s)    continue;
            if(tag)    printf(" ");
            tag=true;
            printf("%d",d[i]);    
        }
        printf("\n");
    }
    return 0;
}