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 = 100000 + 100;
const int MAXM = 110000;
const double eps = 1e-8;
LL MOD = 1000000007;

const LL H = ((LL)1 << 33) - 1;
int t, n, m, num[40], cas = 0;
int tree[3400000][2], s;

void get_num(LL x) {
    mem0(num);
    int id = 0;
    while(x) {
        num[id++] = x % 2;
        x >>= 1;
    }
}

void insert_to_tree(LL x)
{
    get_num(x);
    int u = 0;
    dec (i, 32, 0) {
        int c = num[i];
        if(!tree[u][c]) {
            tree[u][c] = ++s;
        }
        u = tree[u][c];
    }
}

int main()
{
//    FIN;
    cin >> t;
    while(t--) {
        s = 0;
        mem0(tree);
        scanf("%d %d", &n, &m);
        LL x, ans;
        rep (i, 1, n) {
            scanf("%lld", &x);
            insert_to_tree(x);
        }
        printf("Case #%d:
", ++cas);
        rep (i, 1, m) {
            scanf("%lld", &x);
            x = (~x) & H;
            get_num(x);
            int u = 0;
            ans = 0;
            dec (j, 32, 0) {
                int c = num[j];
                ans |= (LL)(tree[u][c] ? c : !c) << j;
                u = tree[u][c] ? tree[u][c] : tree[u][!c];
            }
            printf("%lld
", ans);
        }
    }
    return 0;
}