题意：
给定一棵无根树，可以添加边使之存在环，添加或删除都需要满足环上异或和为0，求最后的最小生成树
做法：
把树通过dfs处理异或和添加边变成一个无向完全图，在这张图上面跑Boruvka算法求最小生成树
Boruvka算法可以解决需要对边之间进行一定处理后的最小生成树问题
CODE
1 #include <bits/stdc++.h> 2 #define dbug(x) cout << #x << "=" << x << endl 3 #define eps 1e-8 4 #define pi acos(-1.0) 5 6 using namespace std; 7 typedef long long LL; 8 9 const int inf = 0x3f3f3f3f; 10 11 template<class T>inline void read(T &res) 12 { 13 char c;T flag=1; 14 while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0'; 15 while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag; 16 } 17 18 const int maxn = 2e5 + 7; 19 20 int n; 21 int edge[maxn << 1], head[maxn << 1], nxt[maxn << 1], len[maxn << 1], cnt; 22 vector<int> a; 23 24 //01字典树 25 struct Tire { 26 int ch[maxn * 31][2], l[maxn * 31], r[maxn * 31]; 27 int sz[maxn * 31]; 28 int siz = 0; 29 LL ans = 0; 30 void insert(int num) { 31 int u = 0; 32 for ( int i = 29; i >= 0; --i ) { 33 int id = (a[num] >> i) & 1; 34 if(!ch[u][id]) { 35 ch[u][id] = ++siz; 36 memset(ch[siz], 0, sizeof(siz)); 37 l[siz] = r[siz] = num; 38 } 39 u = ch[u][id], ++sz[u], l[u] = min(l[u], num), r[u] = max(r[u], num); 40 } 41 } 42 43 int find(int u, int dep, int val) { 44 if(!dep || l[u] == r[u]) { 45 return a[l[u]]; 46 } 47 int id = (val >> (dep - 1)) & 1; 48 if(ch[u][id]) { 49 return find(ch[u][id], dep - 1, val); 50 } 51 else { 52 return find(ch[u][id ^ 1], dep - 1, val); 53 } 54 } 55 56 void dfs(int u, int dep) { 57 if(!dep) { 58 return; 59 } 60 int ls = ch[u][0], rs = ch[u][1]; 61 if(ls && rs) { 62 LL res = 1e17; 63 if(sz[ls] <= sz[rs]) { 64 for ( int i = l[ls]; i <= r[ls]; ++i ) { 65 res = min(res, 1ll * a[i] ^ find(rs, dep - 1, a[i])); 66 } 67 } 68 else { 69 for ( int i = l[rs]; i <= r[rs]; ++i ) { 70 res = min(res, 1ll * a[i] ^ find(ls, dep - 1, a[i])); 71 } 72 } 73 ans += res; 74 } 75 if(sz[ls] > 1) { 76 dfs(ls, dep - 1); 77 } 78 if(sz[rs] > 1) { 79 dfs(rs, dep - 1); 80 } 81 } 82 83 void init() { 84 memset(sz, 0, sizeof(sz)); 85 memset(ch[0], 0, sizeof(ch[0])); 86 siz = ans = 0; 87 } 88 } trie; 89 //01字典树 90 91 void BuildGraph(int u, int v, int w) { 92 ++cnt; 93 edge[cnt] = v; 94 nxt[cnt] = head[u]; 95 head[u] = cnt; 96 len[cnt] = w; 97 } 98 99 void dfs(int u, int fa, int val) { 100 a.push_back(val); 101 for ( int i = head[u]; i; i = nxt[i] ) { 102 int v = edge[i]; 103 if(v == fa) { 104 continue; 105 } 106 dfs(v, u, val ^ len[i]); 107 } 108 } 109 int main() 110 { 111 read(n); 112 int u, v, w; 113 for ( int i = 1; i <= n - 1; ++i ) { 114 read(u); read(v); read(w); 115 ++u, ++v; 116 BuildGraph(u, v, w); 117 BuildGraph(v, u, w); 118 } 119 dfs(1, 0, 0); 120 sort(a.begin(), a.end()); 121 a.erase(unique(a.begin(), a.end()), a.end()); 122 trie.init(); 123 int len = a.size(); 124 for ( int i = 0; i <= len - 1; ++i ) { 125 trie.insert(i); 126 } 127 trie.dfs(0, 30); 128 cout << trie.ans << endl; 129 return 0; 130 }