题目链接

分析：这叫补图上的BFS，萌新第一次遇到= =。方法很简单，看了别人的代码后，自己也学会了。方法就是开两个集合，一个A表示在下一次bfs中能够到达的点，另一个B就是下一次bfs中到不了的点。一开始先把出了起点的所有点都加入A，然后从bfs的点跑一遍边， 把边相连的点从A中取出放到B中。然后遍历A集合，进行bfs。然后把B全部放入A中，清空B。于是又回到了通过边把边连接的点从A移动到B，重复bfs。。。解释地不是很清楚，大体意思就是这样的了。在这个方法中，每个点和每条边都值操作过一次，复杂度是O(N+M)但是set好像有一点常数(？)萌新表示不是很清楚。

代码：

/*****************************************************/

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <map>

#include <set>

#include <ctime>

#include <stack>

#include <queue>

#include <cmath>

#include <string>

#include <vector>

#include <cstdio>

#include <cctype>

#include <cstring>

#include <sstream>

#include <cstdlib>

#include <iostream>

#include <algorithm>

using namespace std;

#define   offcin        ios::sync_with_stdio(false)

#define   sigma_size    26

#define   lson          l,m,v<<1

#define   rson          m+1,r,v<<1|1

#define   slch          v<<1

#define   srch          v<<1|1

#define   sgetmid       int m = (l+r)>>1

#define   LL            long long

#define   ull           unsigned long long

#define   mem(x,v)      memset(x,v,sizeof(x))

#define   lowbit(x)     (x&-x)

#define   bits(a)       __builtin_popcount(a)

#define   mk            make_pair

#define   pb            push_back

#define   fi            first

#define   se            second

const int    INF    = 0x3f3f3f3f;

const LL     INFF   = 1e18;

const double pi     = acos(-1.0);

const double inf    = 1e18;

const double eps    = 1e-9;

const LL     mod    = 1e9+7;

const int    maxmat = 10;

const ull    BASE   = 31;

/*****************************************************/

const int maxn = 2e5 + 5;

std::vector<int> G[maxn];

set<int> in, out;

int N, M;

int dis[maxn];

bool inn[maxn];

void bfs(int s) {

    queue<int> que;

    in.clear(); out.clear();

    mem(inn, false);

    mem(dis, INF);

    for (int i = 1; i <= N; i ++) if (i != s) {

        in.insert(i);

        inn[i] = true;

    }

    dis[s] = 0;

    que.push(s);

    while (!que.empty() && in.size() != 0) {

        int u = que.front(); que.pop();

        for (int i = 0; i < G[u].size(); i ++) {

            int v = G[u][i];

            if (inn[v]) {

                inn[v] = false;

                in.erase(v);

                out.insert(v);

            }

        }

        for (set<int> :: iterator it = in.begin(); it != in.end(); it ++) {

            int v = *it;

            inn[v] = false;

            if (dis[v] > dis[u] + 1) {

                dis[v] = dis[u] + 1;

                que.push(v);

            }

        }

        in.clear();

        for (set<int> :: iterator it = out.begin(); it != out.end(); it ++) {

            int k = *it;

            inn[k] = true;

            in.insert(k);

        }

        out.clear();

    }

}

int main(int argc, char const *argv[]) {

    int T; cin>>T;

    while (T --) {

        scanf("%d%d", &N, &M);

        for (int i = 1; i <= N; i ++) G[i].clear();

        for (int i = 0; i < M; i ++) {

            int u, v;

            scanf("%d%d", &u, &v);

            G[u].pb(v);

            G[v].pb(u);

        }

        int s; scanf("%d", &s);

        bfs(s);

        for (int i = 1; i <= N; i ++) if (i != s) {

            printf("%d", dis[i] == INF ? -1 : dis[i]);

            if (i == N) puts("");

            else printf(" ");

        }

    }

    return 0;

}