[kruskal][Trie] Codeforces 888G Xor-MST
题目描述
You are given a complete undirected graph with nn vertices. A number a_{i}ai is assigned to each vertex, and the weight of an edge between vertices ii and jj is equal to a_{i}xora_{j}aixoraj .
Calculate the weight of the minimum spanning tree in this graph.
输入输出格式
输入格式:
The first line contains nn ( 1<=n<=2000001<=n<=200000 ) — the number of vertices in the graph.
The second line contains nn integers a_{1}a1 , a_{2}a2 , ..., a_{n}an ( 0<=a_{i}<2^{30}0<=ai<230 ) — the numbers assigned to the vertices.
输出格式:
Print one number — the weight of the minimum spanning tree in the graph.
输入输出样例
题解
- 先考虑如何处理异或,比较常规的做法就是建一棵Trie,求最小生成树,则用Kruskal
- 考虑dfs到一个点x,如果它有左右两个儿子,注意左右两子树已经递归地被连通成一个连通块了,那么还需要一条边连接它的左右两棵子树对应的连通块
- 我们可以暴力地找这条边,同时从左右两边走下去,如果能同时走左边就同时走左边,能同时走右边就同时走右边,都能就都走
- 这样的话时间似乎会超时,但根据启发式合并的思想,每次连边的复杂度等于 树的深度×小的子树的大小,故DFS+连边总时间不超过O(nlog^2n)
代码
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #define ll long long 5 using namespace std; 6 int const N=30,M=2e5+1; 7 struct tree{ int son[2],d; }t[N*M]; 8 int n; 9 ll ans,root,tot,r,mi[N],a[M]; 10 void merge(int x,int y,ll k) 11 { 12 if (t[x].d==N) { r=min(r,k); return; } 13 bool flag=false; 14 for (int i=0;i<=1;i++) if (t[x].son[i]&&t[y].son[i]) merge(t[x].son[i],t[y].son[i],k),flag=true; 15 if (!flag) 16 { 17 if (t[x].son[0]&&t[y].son[1]) merge(t[x].son[0],t[y].son[1],k+mi[N-t[x].d-1]); else merge(t[x].son[1],t[y].son[0],k+mi[N-t[x].d-1]); 18 } 19 } 20 void dfs(int x) 21 { 22 if (!x||t[x].d==N) return; 23 for (int i=0;i<=1;i++) dfs(t[x].son[i]); 24 if (t[x].son[0]&&t[x].son[1]) r=2e9,merge(t[x].son[0],t[x].son[1],mi[N-t[x].d-1]),ans+=r; 25 } 26 int main() 27 { 28 mi[0]=1; for (int i=1;i<N;i++) mi[i]=mi[i-1]*2; 29 scanf("%d",&n),root=tot=1; 30 for (int i=1,x;i<=n;i++) 31 { 32 scanf("%lld",&a[i]),x=root; 33 for (int j=0;j<N;j++) 34 { 35 int r=((mi[N-j-1]&a[i])>0); 36 if (!t[x].son[r]) t[x].son[r]=++tot,t[tot].d=t[x].d+1; 37 x=t[x].son[r]; 38 } 39 } 40 dfs(1),printf("%lld",ans); 41 }