[XOR minimum spanning tree divide-and-conquer dictionary tree merge] BNUOJ52318 be Friends

Binary from high to Low Division, current division to the first I-bit

Then it must be put the first I i bit 0 together, for 1 of the put together, the two pieces into a unicom block, and then a side is the best

Two pieces of the rule go down.

Look for the smallest side, build a dictionary tree on the other side, and look for the number on the dictionary tree.

#include <cstdio> #include <iostream> #include <algorithm> using namespace std;

const int n=100010;

typedef long Long LL;
int n,cnt,a[n],nxt[n*50][2];

ll ans; void Add (int cur,int x) {for (int i=30;~i;i--) {if (!nxt[cur][x>>i&1]) NXT[CUR][X&GT;&GT;I&AMP;1]=++CN
        T
    cur=nxt[cur][x>>i&1];
    }} int Query (int cur,int x) {int ret=0;
        for (int i=30;~i;i--) {if (nxt[cur][x>>i&1]) cur=nxt[cur][x>>i&1];
    else ret+=1<<i,cur=nxt[cur][~x>>i&1];
} return ret;

} int LC;

inline BOOL CMP (const int &AMP;A,CONST int &b) {return (a>>lc&1) < (b>>lc&1);}
    int Merge (int x,int y) {if (!x | |!y) return x+y;
    Nxt[x][0]=merge (Nxt[x][0],nxt[y][0]);
    Nxt[x][1]=merge (nxt[x][1],nxt[y][1]);
return x; } int solve (int l,int r,int t) {if (!t) {int ro=++cnt;
        ADD (Ro,a[l]);
    return ro; } lc=t-1;
    Sort (a+l,a+1+r,cmp); int miD=L-1; for (int i=l;i<=r;i++) if (a[i]>>lc&1) break;
    else mid=i;
    int lro=0,rro=0,cur=1<<30;
    if (l<=mid) lro=solve (l,mid,t-1);
    if (mid+1<=r) rro=solve (mid+1,r,t-1);
    for (int i=l;i<=mid;i++) cur=min (Cur,query (Rro,a[i]));
    if (mid<l | | mid>=r) cur=0;
    Ans+=cur;
Return Merge (Lro,rro);
    } int main () {scanf ("%d", &n);
    for (int i=1;i<=n;i++) scanf ("%d", &a[i]);
    Solve (1,n,31);
    cout<<ans<<endl;
return 0;  }