Codeforces Round #536 (Div. 2)

Problem  Codeforces Round #536 (Div. 2) - D. Lunar New Year and a Wander

Time Limit: 3000 mSec

Problem Description

Input

Output

Output a line containing the lexicographically smallest sequence a1,a2,…,an Bob can record.

Sample Input

3 2
1 2
1 3

Sample Output

1 2 3

题解：这一场的D和E都是在经典模型的基础上稍加改动，题目很有质量。根据题意就可以想出第一个最暴力的算法，遍历已经遍历到的点，找其相邻节点中未被遍历到的最小的点，这个正确性肯定是没问题的，但是复杂度是O(n^2)，再一想，这些已经遍历到的点肯定是组成了一个联通块，并查集维护一下，每次合并的时候把加入集合的点的相邻节点遍历一下，加入到优先队列中不就行了，这不就是Dijkstra算法的思路么，这里的距离是不在集合中的点到集合的距离，按字典序大小给出距离，加入集合的顺序即为答案，轻松搞定。

 

#include <bits/stdc++.h>

using namespace std;

#define REP(i, n) for (int i = 1; i <= (n); i++)
#define sqr(x) ((x) * (x))

const int maxn = 100000 + 100;
const int maxm = 200000 + 100;
const int maxs = 256;

typedef long long LL;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;

const LL unit = 1LL;
const int INF = 0x3f3f3f3f;
const double eps = 1e-14;
const double inf = 1e15;
const double pi = acos(-1.0);
const int SIZE = 100 + 5;
const LL MOD = 1000000007;

struct Edge
{
    int to, next;
} edge[maxm<<1];

struct HeapNode
{
    int u;
    bool operator< (const HeapNode &a)const
    {
        return u > a.u;
    }
};

int n, m;
int head[maxn], tot;
int dist[maxn];
vector<int> ans;
bool vis[maxn];

void init()
{
    memset(head, -1, sizeof(head));
    tot = 0;
}

void AddEdge(int u, int v)
{
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}

void Dijkstra()
{
    memset(dist, INF, sizeof(dist));
    dist[0] = 0;
    priority_queue<HeapNode> que;
    que.push((HeapNode{0}));
    while(!que.empty())
    {
        HeapNode x = que.top();
        int u = x.u;
        que.pop();
        vis[u] = true;
        ans.push_back(u + 1);
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
            if(vis[v])
                continue;
            if(dist[v] == INF)
            {
                dist[v] = v;
                que.push((HeapNode){v});
            }
        }
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    cin >> n >> m;
    init();
    int u, v;
    for(int i = 0; i < m; i++)
    {
        cin >> u >> v;
        u--, v--;
        AddEdge(u, v);
        AddEdge(v, u);
    }
    Dijkstra();
    for(int i = 0; i < ans.size(); i++)
    {
        cout << ans[i] << " ";
    }
    return 0;
}