CF #541div2 F

题目本质：并查集的链式合并

解决方法1:

类似哈夫曼树，叶节点们为真点，其余造一些虚的父节点，使得dfs这棵树的时候，先进行并查合并的点一定是兄弟节点因而紧挨着被输出，巧妙达到了效果。

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <ctime>
 #include <cctype>
 #include <climits>
 #include <iostream>
 #include <iomanip>
 #include <algorithm>
 #include <string>
 #include <sstream>
 #include <stack>
 #include <queue>
 #include <set>
 #include <map>
 #include <vector>
 #include <list>
 #include <fstream>
 #define ri readint()
 #define gc getchar()
 #define R(x) scanf("%d", &x)
 #define W(x) printf("%d\n", x)
 #define init(a, b) memset(a, b, sizeof(a))
 #define rep(i, a, b) for (int i = a; i <= b; i++)
 #define irep(i, a, b) for (int i = a; i >= b; i--)
 #define ls p << 1
 #define rs p << 1 | 1
 using namespace std;

 typedef double db;
 typedef long long ll;
 typedef unsigned long long ull;
 typedef pair<int, int> P;
 const int inf = 0x3f3f3f3f;
 const ll INF = 1e18;

 inline int readint() {
     int x = , s = , c = gc;
     while (c <= )    c = gc;
     if (c == '-')    s = -, c = gc;
     for (; isdigit(c); c = gc)
         x = x *  + c - ;
     return x * s;
 }

 const int maxn = 15e4 + ;
 int n, fa[maxn << ], l[maxn << ], r[maxn << ];

 inline int find(int v) {
     return v == fa[v] ? v : fa[v] = find(fa[v]);
 }

 void dfs(int cur) {
     if (!l[cur] && !r[cur]){
         printf("%d ", cur);
         return;
     }
     if (l[cur])    dfs(l[cur]);
     if (r[cur])    dfs(r[cur]);
 }

 int main() {
     n = ri;
     rep(i, , n) {
         fa[i] = i;
     }
     rep(i, , n - ) {
         int a = ri, b = ri;
         a = find(a), b = find(b);
         fa[a] = fa[b] = fa[n + i] = n + i;
         l[n + i] = a, r[n + i] = b;
     }
     dfs(n + n - );
     return ;
 }

解决方法2:

正常地用数组记录链，l和r记录真实的左右顺序，并查集式的getl和getr记录这个链上的最左端和最右端，两个集合并时一接。貌似直接把常规并查集的father数组扔了……

大佬代码：

 int n,m,i,j,k,a[],l[],r[],fl[],fr[],x,y;
 int gfl(int x){if (fl[x]==x) return x;return fl[x]=gfl(fl[x]);}
 int gfr(int x){if (fr[x]==x) return x;return fr[x]=gfr(fr[x]);}
 int main()
 {
     read(n);
     rep(i,n)
     {
         fl[i]=fr[i]=l[i]=r[i]=i;
     }
     rep(i,n-)
     {
         read(x);read(y);
         x=gfr(x);y=gfl(y);
         r[x]=y;l[y]=x;
         fr[x]=y;fl[y]=x;
     }
     x=gfl();
     rep(i,n)
     {
         printf("%d ",x);
         x=r[x];
     }
     return ;
 }