hdu5876——Sparse Graph（补图的最短路）

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

题意：给出一个原图，求补图的最短路
分析：比赛的时候一心想先建完全图，再忽略原图中的边来求最短路，换了各种模板，结果不是时间超就是内存超，结论是不能建完全图来算这道题（二十万个点的完全图。。。），结果比赛完了也没想到方法，我真的是很菜。。。
网上找到的题解是存原图，暴力枚举所有点，如果与起点的边不在原图中，那么距离就是起点的距离+1，然后用BFS再把这些点设为起点，再枚举剩下的没算出距离的点。。。
原博主写得有点抽象，我照着自己的理解写了份代码

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#include <set>
#include <cmath>
#include <algorithm>
#define INF 0x3f3f3f3f
#define MAXN 200005
#define Mod 10001
using namespace std;
int n,m,dis[MAXN];
set<int> all,mp[MAXN],tmp;
queue<int> q;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;++i)
        {
            all.insert(i);  //先把所有顶点加入
            mp[i].clear();
        }
        memset(dis,-1,sizeof(dis));
        int u,v,s;
        while(m--)
        {
            scanf("%d%d",&u,&v);
            mp[u].insert(v);
            mp[v].insert(u);
        }
        scanf("%d",&s);
        q.push(s);
        dis[s]=0;
        all.erase(s);
        while(!q.empty())
        {
            tmp.clear();
            u=q.front();
            q.pop();
            set<int>::iterator v=all.begin();
            for(;v!=all.end();v++)  //遍历所有顶点
            {
                if(mp[u].find(*v)==mp[u].end()) //不在原图中的边说明补图中可以到达
                {
                    dis[*v]=dis[u]+1;
                    tmp.insert(*v);  //这些点下次就不用考虑
                    q.push(*v); //以当前距离为起点（第一次距离为1），寻找距离+1的点
                }
            }
            for(v=tmp.begin();v!=tmp.end();v++) //在所有顶点中删除已经算过的点
                all.erase(*v);
        }
        bool flag=false;
        for(int i=1;i<=n;++i)
        {
            if(i!=s)
            {
                if(flag)
                    printf(" %d",dis[i]);
                else
                    printf("%d",dis[i]);
                flag=true;
            }
        }
        printf("\n");
    }
    return 0;
}