HDU 4825 字典树
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）
1 //#pragma comment(linker, "/STACK:1677721600") 2 #include <map> 3 #include <set> 4 #include <stack> 5 #include <queue> 6 #include <cmath> 7 #include <ctime> 8 #include <vector> 9 #include <cstdio> 10 #include <cctype> 11 #include <cstring> 12 #include <cstdlib> 13 #include <iostream> 14 #include <algorithm> 15 using namespace std; 16 #define INF 0x3f3f3f3f 17 #define inf (-((LL)1<<40)) 18 #define lson k<<1, L, (L + R)>>1 19 #define rson k<<1|1, ((L + R)>>1) + 1, R 20 #define mem0(a) memset(a,0,sizeof(a)) 21 #define mem1(a) memset(a,-1,sizeof(a)) 22 #define mem(a, b) memset(a, b, sizeof(a)) 23 #define FIN freopen("in.txt", "r", stdin) 24 #define FOUT freopen("out.txt", "w", stdout) 25 #define rep(i, a, b) for(int i = a; i <= b; i ++) 26 #define dec(i, a, b) for(int i = a; i >= b; i --) 27 28 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 29 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 30 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 31 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 32 33 //typedef __int64 LL; 34 typedef long long LL; 35 const int MAXN = 100000 + 100; 36 const int MAXM = 110000; 37 const double eps = 1e-8; 38 LL MOD = 1000000007; 39 40 const LL H = ((LL)1 << 33) - 1; 41 int t, n, m, num[40], cas = 0; 42 int tree[3400000][2], s; 43 44 void get_num(LL x) { 45 mem0(num); 46 int id = 0; 47 while(x) { 48 num[id++] = x % 2; 49 x >>= 1; 50 } 51 } 52 53 void insert_to_tree(LL x) 54 { 55 get_num(x); 56 int u = 0; 57 dec (i, 32, 0) { 58 int c = num[i]; 59 if(!tree[u][c]) { 60 tree[u][c] = ++s; 61 } 62 u = tree[u][c]; 63 } 64 } 65 66 int main() 67 { 68 // FIN; 69 cin >> t; 70 while(t--) { 71 s = 0; 72 mem0(tree); 73 scanf("%d %d", &n, &m); 74 LL x, ans; 75 rep (i, 1, n) { 76 scanf("%lld", &x); 77 insert_to_tree(x); 78 } 79 printf("Case #%d:\n", ++cas); 80 rep (i, 1, m) { 81 scanf("%lld", &x); 82 x = (~x) & H; 83 get_num(x); 84 int u = 0; 85 ans = 0; 86 dec (j, 32, 0) { 87 int c = num[j]; 88 ans |= (LL)(tree[u][c] ? c : !c) << j; 89 u = tree[u][c] ? tree[u][c] : tree[u][!c]; 90 } 91 printf("%lld\n", ans); 92 } 93 } 94 return 0; 95 }
