HDU-4825 Xor Sum &&Codeforces Round #367 (Div. 2) D. Vasiliy's Multiset (Trie树)

来源：互联网发布：中国港口数据编辑：程序博客网时间：2024/05/05 11:33

Xor Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 1427 Accepted Submission(s): 595

Problem Description

Zeus 和 Prometheus 做了一个游戏，Prometheus 给 Zeus 一个集合，集合中包含了N个正整数，随后 Prometheus 将向 Zeus 发起M次询问，每次询问中包含一个正整数 S ，之后 Zeus 需要在集合当中找出一个正整数 K ，使得 K 与 S 的异或结果最大。Prometheus 为了让 Zeus 看到人类的伟大，随即同意 Zeus 可以向人类求助。你能证明人类的智慧么？

Input

输入包含若干组测试数据，每组测试数据包含若干行。
输入的第一行是一个整数T（T < 10），表示共有T组数据。
每组数据的第一行输入两个正整数N，M（<1=N,M<=100000），接下来一行，包含N个正整数，代表 Zeus 的获得的集合，之后M行，每行一个正整数S，代表 Prometheus 询问的正整数。所有正整数均不超过2^32。

Output

对于每组数据，首先需要输出单独一行”Case #?:”，其中问号处应填入当前的数据组数，组数从1开始计算。
对于每个询问，输出一个正整数K，使得K与S异或值最大。

Sample Input

23 23 4 5154 14 6 5 63

Sample Output

Case #1:43Case #2:4

Source

2014年百度之星程序设计大赛 - 资格赛

对Trie树理解不到位，照着jhz033的套路写了

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 2000005;int a[N][3];int sz;void Init(){    memset(a,0,sizeof(a));    sz = 1;}void Insert(int x){    int num[35];    memset(num,0,sizeof(num));    int cnt = 0;    while(x)    {        num[cnt++] = x % 2;        x /= 2;    }    int u = 0,n = 34;    for(int i = n;i >= 0;i--)    {        if(!a[u][num[i]])            a[u][num[i]] = sz++;        u = a[u][num[i]];    }}int Query(int y){    int num[35];    memset(num,0,sizeof(num));    int cnt = 0;    while(y)    {        num[cnt++] = y % 2;        y /= 2;    }    for(int i = 0;i <= 34;i++)        num[i]= num[i] ? 0 : 1;    int u = 0,n = 34,v,w;    int ans = 0;    for(int i = n;i >= 0;i--)    {        if(num[i])        {            v = 1;            w = 0;        }        else        {            w = 1;            v = 0;        }        if(a[u][v])        {            u =  a[u][v];            if(v)                ans += 1 << i;        }        else        {            u = a[u][w];            if(!v)                ans += 1 << i;        }    }    return ans;}int main(){    int t,n,m;    cin >> t;    for(int ca = 1;ca <= t;ca++)    {        Init();        cin >> n >> m;        int x,y;        for(int i = 1;i <= n;i++)        {            scanf("%d",&x);            Insert(x);        }        printf("Case #%d:\n",ca);        for(int i = 1;i <= m;i++)        {            scanf("%d",&y);            printf("%d\n",Query(y));        }    }    return 0;}

Codeforces Round #367 (Div. 2)

Author has gone out of the stories about Vasiliy, so here is just a formal task description.

You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:

"+ x" — add integer x to multiset A.
"- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query.
"? x" — you are given integer x and need to compute the value $\text{[math]}$ , i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.

Multiset is a set, where equal elements are allowed.

Input

The first line of the input contains a single integer q (1 ≤ q ≤ 200 000) — the number of queries Vasiliy has to perform.

Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer x_i (1 ≤ x_i ≤ 10⁹). It's guaranteed that there is at least one query of the third type.

Note, that the integer 0 will always be present in the set A.

Output

For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer x_i and some integer from the multiset A.

Example

input

10+ 8+ 9+ 11+ 6+ 1? 3- 8? 3? 8? 11

output

11101413

Note

After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.

The answer for the sixth query is integer $\text{[math]}$ — maximum among integers $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ and $\text{[math]}$ .

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 4000005;int a[N][3],sum[N];int sz;void Init(){    memset(a,0,sizeof(a));    memset(sum,0,sizeof(sum));    sz = 1;}void Insert(int x){    int num[35];    memset(num,0,sizeof(num));    int cnt = 0;    while(x)    {        num[cnt++] = x % 2;        x /= 2;    }    int u = 0,n = 34;    for(int i = n;i >= 0;i--)    {        if(!a[u][num[i]])            a[u][num[i]] = sz++;        u = a[u][num[i]];        sum[u]++;    }}void Delete(int x){    int num[35];    memset(num,0,sizeof(num));    int cnt = 0;    while(x)    {        num[cnt++] = x % 2;        x /= 2;    }    int u = 0,n = 34;    for(int i = n;i >= 0;i--)    {        int v = a[u][num[i]];        sum[v]--;        if(!sum[v])            a[u][num[i]] = 0;        u = v;    }}int Query(int y){    int num[35];    memset(num,0,sizeof(num));    int cnt = 0;    while(y)    {        num[cnt++] = y % 2;        y /= 2;    }    for(int i = 0;i <= 34;i++)        num[i]= num[i] ? 0 : 1;    int u = 0,n = 34,v,w;    int ans = 0;    for(int i = n;i >= 0;i--)    {        if(num[i])        {            v = 1;            w = 0;        }        else        {            w = 1;            v = 0;        }        if(a[u][v])        {            u =  a[u][v];            ans += 1 << i;        }        else        {            u = a[u][w];            //if(!v)                //ans += 1 << i;        }    }    return ans;}int main(){    int t,x;    cin >> t;    char s[20];    Init();    Insert(0);    for(int ca = 1;ca <= t;ca++)    {        scanf("%s%d",s,&x);        if(s[0] == '+')            Insert(x);        else if(s[0] == '-')            Delete(x);        else            printf("%d\n",Query(x));    }    return 0;}

感觉这样的代码有点丑陋，但是原理应该一样，弄熟了回来改

0 0