HDU 5876 大连网络赛 Sparse Graph

Sparse Graph

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 928    Accepted Submission(s): 312

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

补图上的BFS
因为补图是一个完全图，然后去掉最多两万条边，有可以想象，从起点到任何一点的最短距离肯定很短，我可以从起点开始扫
把所有没有和起点有边的（原图中）入队列，这是第一层距离为1的，然后以这些点继续扩展，因为最短距离很短，最多扩展10
次就把所有点都扫过了，已经扩展过的点在bfs过程中就不要扫了，所以可以set或者vector把扩展过的点删除
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <set>
#include <vector>
#include <queue>

using namespace std;
const int maxn=2*1e5;
int n,m,k;
int d[maxn+5];
set<int> a[maxn+5];
set<int> s;
queue<int> q;
int de[maxn+5];
void BFS(int x)
{
    q.push(x);
    memset(d,-1,sizeof(d));
    d[x]=0;
    while(!q.empty())
    {
        int term=q.front();
        q.pop();
        int cnt=0;
        for(set<int>::iterator it=s.begin();it!=s.end();it++)
        {
            bool tag=true;
            int i=*it;
            if(a[i].find(term)==a[i].end())
            {
                d[i]=d[term]+1;
                q.push(i);
                de[cnt++]=i;
            }
        }
        for(int i=0;i<cnt;i++)
            s.erase(de[i]);
    }
}
int main()
{
    int t;
    int x,y;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        s.clear();
        for(int i=1;i<=n;i++)
            s.insert(i),a[i].clear();
        for(int i=1;i<=m;i++)
        {
             scanf("%d%d",&x,&y);
             a[x].insert(y);
             a[y].insert(x);
        }

        scanf("%d",&k);
        s.erase(k);
        BFS(k);
        for(int i=1;i<=n;i++)
        {
            if(i==k)
                continue;
            printf("%d",d[i]);
            if(i!=n)
                printf(" ");
        }
        printf("\n");
    }
    return 0;
}