题目链接

Solution

\(Trie\) 树 + 启发式合并.
考虑到是异或,于是按位贪心.让高位的尽量相同.
然后要计算每棵子树的代价,似乎并没有很好的方法??
于是只能启发式合并.
对于每一个有两个子节点的点;
将 \(siz\) 较小的点中的值放到 \(siz\) 较大的子树中去查询即可.
时间复杂度 \(O(n(logn)^2)\) .

Code

Dark 鸡哥 的代码:

//It is made by Awson on 2018.3.18
#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int N = 200000;
const LL INF = 1e18;int n, a[N+5], A[40]; LL bin[40], ans;
struct Trie {int ch[N*35][2], bit[N*35+5], pos;vector<int>key[N*35];void insert(int t) {int u = 0; key[u].push_back(t);for (int i = 35; i >= 1; i--) {if (ch[u][A[i]] == 0) ch[u][A[i]] = ++pos; u = ch[u][A[i]];t -= bin[i-1]*A[i], key[u].push_back(t), bit[u] = i;}}LL qry(int u, int t) {LL ans = 0;for (int i = 1; i <= bit[u]-1; i++) A[i] = t%2, t /= 2;for (int i = bit[u]-1; i >= 1; i--) {if (ch[u][A[i]] != 0) u = ch[u][A[i]];else u = ch[u][A[i]^1], ans += bin[bit[u]-1];}return ans;}LL query(int o) {if (key[o].size() == 1) return 0;if (ch[o][0]) ans += query(ch[o][0]);if (ch[o][1]) ans += query(ch[o][1]);if (!ch[o][0] || !ch[o][1]) return 0;LL tmp = INF; int flag = 0;if (key[ch[o][0]].size() > key[ch[o][1]].size()) flag = 1;int sz = key[ch[o][flag]].size();for (int i = 0; i < sz; i++) tmp = min(qry(ch[o][flag^1], key[ch[o][flag]][i]), tmp);return tmp+bin[bit[o]-2];}
}T;void work() {scanf("%d", &n); bin[0] = 1; for (int i = 1; i <= 35; i++) bin[i] = (bin[i-1]<<1);for (int i = 1; i <= n; i++) scanf("%d", &a[i]);sort(a+1, a+n+1); n = unique(a+1, a+n+1)-a-1;for (int i = 1; i <= n; i++) {for (int j = 1, t = a[i]; j <= 35; j++) A[j] = t%2, t /= 2; T.insert(a[i]);}ans += T.query(0); printf("%I64d\n", ans);
}
int main() {work(); return 0; }

我的代码(有问题):

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll maxn=250008;
const ll inf=1926081737;ll ch[maxn*40][2];
ll n,tot,t[maxn*40];
ll w[maxn],cf[43],ans;
vector <ll>a[maxn];void insert(ll x)
{ll u=0;for(ll i=35;i>=0;i--){ll c=(x>>i)&1;if(!ch[u][c]) ch[u][c]=++tot;a[u].push_back(x);x-=c*cf[i];t[u]=i;u=ch0[u][c];}
}ll query(ll x,ll u)
{ll now=0,nn=t[u];for(ll i=nn;i>=0;i--){ll c=(x>>i)&1;if(ch[u][c]) u=ch[u][c];else u=ch[u][c^1],now+=cf[i];}return now;
}void solve(ll u)
{if(a[u].size()==1) return;for(ll i=0;i<2;i++)if(ch[u][i]) solve(ch[u][i]);if(!ch[u][0]||!ch[u][1]) return;ll lx=ch[u][0],rx=ch[u][1];if(a[lx].size()>a[rx].size())swap(lx,rx);ll now=inf,nn=a[lx].size();for(ll i=0;i<nn;i++)now=min(now,query(a[lx][i],rx));now=(now==inf?0:now);ans+=cf[t[u]]+now;return ;
}int main()
{freopen("a.in","r",stdin);freopen("a.ans","w",stdout);scanf("%lld",&n);cf[0]=1; for(ll i=1;i<=40;i++) cf[i]=cf[i-1]*2;for(ll i=1;i<=n;i++)scanf("%lld",&w[i]);sort(w+1,w+n+1);  n=unique(w+1,w+n+1)-w-1;for(ll i=1;i<=n;i++) insert(w[i]);solve(0);printf("%lld\n",ans);return 0;
}