[CF888G] Xor-mst (Trie 树,最小生成树)

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 
#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;
    vectorkey[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
#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 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

转载于:https://www.cnblogs.com/Kv-Stalin/p/9588612.html