给2个长度为n的数组a, p, 改变p顺序,使得a^p 得出来的数组字典序最小
思路: 以p数组建字典树, 然后每次找与ai最“接近”的数(即俩数异或最小, 每次再从字典树中删除该数即可
#include "iostream" #include "iomanip" #include "string.h" #include "stack" #include "queue" #include "string" #include "vector" #include "set" #include "map" #include "algorithm" #include "stdio.h" #include "math.h" #pragma comment(linker, "/STACK:102400000,102400000") #define bug(x) cout<<x<<" "<<"UUUUU"<<endl; #define mem(a,x) memset(a,x,sizeof(a)) #define step(x) fixed<< setprecision(x)<< #define mp(x,y) make_pair(x,y) #define pb(x) push_back(x) #define ll long long #define endl (" ") #define ft first #define sd second #define lrt (rt<<1) #define rrt (rt<<1|1) using namespace std; const ll mod=1e9+7; const ll INF = 1e18+1LL; const int inf = 1e9+1e8; const double PI=acos(-1.0); const int N=1e5+100; int tree[N*50][2], data[N*50][2]; int a[N*3], p[N*3], n, tot=1, root=1; void Insert(int x) { int now = root, num[31]; for(int i=30; i>=0; --i) { num[i] = x&1; x >>= 1; } for(int i=0; i<=30; ++i) { int t = num[i];//cout<<t<<endl; if(tree[now][t]) data[now][t]++; else{ tree[now][t] = ++tot; data[now][t] = 1; } now = tree[now][t]; } } void del(int x) { int now = root, num[31]; for(int i=30; i>=0; --i) { num[i] = x&1; x >>= 1; } for(int i=0; i<=30; ++i) { int t = num[i]; data[now][t]--; now = tree[now][t]; if(!data[now][t]) tree[now][t] = 0; } } int finds(int x) { int now = root, r = 1, ret = 0; int num[31]; for(int i=30; i>=0; --i) { num[i] = x&1; x >>= 1; } for(int i=0; i<=30; ++i) { int t = num[i]; if(data[now][t]) { ret *= 2; ret += t; now = tree[now][t]; } else { ret *= 2; ret += t^1; now = tree[now][t^1]; } } return ret; } int main() { scanf("%d", &n); for(int i=1; i<=n; ++i) { scanf("%d", &a[i]); } for(int i=1; i<=n; ++i) { scanf("%d", &p[i]); Insert(p[i]); }//cout<<tot<<endl; for(int i=1; i<=n; ++i){ int x = finds(a[i]); del(x); printf("%d ", x^a[i]); } return 0; }