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

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。

  1 #include <bits/stdc++.h>
  2
  3 using namespace std;
  4
  5 #define REP(i, n) for (int i = 1; i <= (n); i++)
  6 #define sqr(x) ((x) * (x))
  7
  8 const int maxn = 100000 + 100;
  9 const int maxm = 200000 + 100;
 10 const int maxs = 256;
 11
 12 typedef long long LL;
 13 typedef pair<int, int> pii;
 14 typedef pair<double, double> pdd;
 15
 16 const LL unit = 1LL;
 17 const int INF = 0x3f3f3f3f;
 18 const double eps = 1e-14;
 19 const double inf = 1e15;
 20 const double pi = acos(-1.0);
 21 const int SIZE = 100 + 5;
 22 const LL MOD = 1000000007;
 23
 24 struct Edge
 25 {
 26     int to, next;
 27 } edge[maxm<<1];
 28
 29 struct HeapNode
 30 {
 31     int u;
 32     bool operator< (const HeapNode &a)const
 33     {
 34         return u > a.u;
 35     }
 36 };
 37
 38 int n, m;
 39 int head[maxn], tot;
 40 int dist[maxn];
 41 vector<int> ans;
 42 bool vis[maxn];
 43
 44 void init()
 45 {
 46     memset(head, -1, sizeof(head));
 47     tot = 0;
 48 }
 49
 50 void AddEdge(int u, int v)
 51 {
 52     edge[tot].to = v;
 53     edge[tot].next = head[u];
 54     head[u] = tot++;
 55 }
 56
 57 void Dijkstra()
 58 {
 59     memset(dist, INF, sizeof(dist));
 60     dist[0] = 0;
 61     priority_queue<HeapNode> que;
 62     que.push((HeapNode{0}));
 63     while(!que.empty())
 64     {
 65         HeapNode x = que.top();
 66         int u = x.u;
 67         que.pop();
 68         vis[u] = true;
 69         ans.push_back(u + 1);
 70         for(int i = head[u]; i != -1; i = edge[i].next)
 71         {
 72             int v = edge[i].to;
 73             if(vis[v])
 74                 continue;
 75             if(dist[v] == INF)
 76             {
 77                 dist[v] = v;
 78                 que.push((HeapNode){v});
 79             }
 80         }
 81     }
 82 }
 83
 84 int main()
 85 {
 86     ios::sync_with_stdio(false);
 87     cin.tie(0);
 88     //freopen("input.txt", "r", stdin);
 89     //freopen("output.txt", "w", stdout);
 90     cin >> n >> m;
 91     init();
 92     int u, v;
 93     for(int i = 0; i < m; i++)
 94     {
 95         cin >> u >> v;
 96         u--, v--;
 97         AddEdge(u, v);
 98         AddEdge(v, u);
 99     }
100     Dijkstra();
101     for(int i = 0; i < ans.size(); i++)
102     {
103         cout << ans[i] << " ";
104     }
105     return 0;
106 }

原文地址：https://www.cnblogs.com/npugen/p/10800220.html