HDU 4825 字典树

HDU 4825

对于给定的查询（一个整数），求集合中和他异或值最大的值是多少

按位从高位往低位建树，查询时先将查询取反，然后从高位往低位在树上匹配，可以匹配不可以匹配都走同一条边（匹配表示有一个异或值为1的边，选择当然最好；不能匹配说明不存在一条异或值为1的边，那么只存在一条为0的边，也不得不选）

 //#pragma comment(linker, "/STACK:1677721600")
 #include <map>
 #include <set>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <ctime>
 #include <vector>
 #include <cstdio>
 #include <cctype>
 #include <cstring>
 #include <cstdlib>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 #define INF 0x3f3f3f3f
 #define inf (-((LL)1<<40))
 #define lson k<<1, L, (L + R)>>1
 #define rson k<<1|1,  ((L + R)>>1) + 1, R
 #define mem0(a) memset(a,0,sizeof(a))
 #define mem1(a) memset(a,-1,sizeof(a))
 #define mem(a, b) memset(a, b, sizeof(a))
 #define FIN freopen("in.txt", "r", stdin)
 #define FOUT freopen("out.txt", "w", stdout)
 #define rep(i, a, b) for(int i = a; i <= b; i ++)
 #define dec(i, a, b) for(int i = a; i >= b; i --)

 template<class T> T MAX(T a, T b) { return a > b ? a : b; }
 template<class T> T MIN(T a, T b) { return a < b ? a : b; }
 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }

 //typedef __int64 LL;
 typedef long long LL;
 const int MAXN =  + ;
 const int MAXM = ;
 const double eps = 1e-;
 LL MOD = ;

 const LL H = ((LL) << ) - ;
 int t, n, m, num[], cas = ;
 int tree[][], s;

 void get_num(LL x) {
     mem0(num);
     int id = ;
     while(x) {
         num[id++] = x % ;
         x >>= ;
     }
 }

 void insert_to_tree(LL x)
 {
     get_num(x);
     int u = ;
     dec (i, , ) {
         int c = num[i];
         if(!tree[u][c]) {
             tree[u][c] = ++s;
         }
         u = tree[u][c];
     }
 }

 int main()
 {
 //    FIN;
     cin >> t;
     while(t--) {
         s = ;
         mem0(tree);
         scanf("%d %d", &n, &m);
         LL x, ans;
         rep (i, , n) {
             scanf("%lld", &x);
             insert_to_tree(x);
         }
         printf("Case #%d:\n", ++cas);
         rep (i, , m) {
             scanf("%lld", &x);
             x = (~x) & H;
             get_num(x);
             int u = ;
             ans = ;
             dec (j, , ) {
                 int c = num[j];
                 ans |= (LL)(tree[u][c] ? c : !c) << j;
                 u = tree[u][c] ? tree[u][c] : tree[u][!c];
             }
             printf("%lld\n", ans);
         }
     }
     return ;
 }