HDU 5876 Sparse Graph BFS+set删点

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

**题意：**给你n个点的完全图，再给你m条边需要删除，给定s，问s到其他所有点的距离 \\(2\le N\le 200000 \\) \\(0\le M\le 20000\\)
**思路：**因为边权值为1，所以直接BFS，再者就是考虑如何枚举点了，用vis标记的话，时间会不够（TLE了一发），在这里使用set维护点集（多方便）但是要注意把点从集合中去除时，不能直接边遍历边去，而是一遍遍历完一起去掉，因为使用的迭代器地址不会随删除而改变...

/** @Date    : 2016-11-14-17.20

  * @Author  : Lweleth (SoungEarlf@gmail.com)

  * @Link    : https://github.com/

  * @Version :

  */

#include <stdio.h>

#include <iostream>

#include <string.h>

#include <algorithm>

#include <utility>

#include <vector>

#include <map>

#include <set>

#include <string>

#include <stack>

#include <queue>

#define pii pair<int , int>

#define MP(x, y) make_pair((x) ,(y))

#define ff first

#define ss second

#define LL long long

#define MMF(x) memset((x),0,sizeof(x))

#define MMI(x) memset((x), INF, sizeof(x))

using namespace std;



const int INF = 0x3f3f3f3f;

const int N = 2e5+20;



map<pii ,bool>mp;

set<int>st;

set<int>::iterator it;

int res[N];

int n, m;

void bfs(int s,int n)

{

    queue<int>q, t;



    res[s] = 0;

    q.push(s);



    st.erase(s);

    while(!q.empty())

    {

        int nw = q.front();

        //cout << nw;

        q.pop();

        for(it = st.begin(); it != st.end(); it++)

        {

            //cout << "~"<< *it << " " ;

            if(!mp[MP(nw, *it)])

            {

                res[*it] = res[nw] + 1;

                q.push(*it);

                t.push(*it);

                //st.erase(*it);//不能直接删会有问题

                //if(st.empty())

                    //break;



            }

        }

        while(!t.empty())

        {

            int x = t.front();

            t.pop();

            st.erase(x);

        }

        //cout << endl;

    }



}



int main()

{



    int T;

    while(~scanf("%d", &T))

    {

        while(T--)

        {

            mp.clear();

            st.clear();

            scanf("%d%d", &n, &m);



            for(int i = 1; i <= n; i++)

                st.insert(i);



            while(m--)

            {

                int x, y;

                scanf("%d%d", &x, &y);

                mp[MP(x, y)] = mp[MP(y, x)] = 1;

            }



            int s;

            scanf("%d", &s);

            bfs(s, n);

            ////

            int flag = 0;

            for(int i = 1; i <= n; i++)

            {

                if(i != s)

                {

                    if(flag)

                        printf(" ");

                    if(res[i] != 0)

                        printf("%d", res[i]);

                    else printf("-1");

                    flag = 1;

                }

            }

            printf("\n");

        }







    }

    return 0;

}

/*

99

6 9

1 3

1 4

2 3

2 4

2 5

3 5

3 6

4 6

5 6

1

*/