hdu5536 Chip Factory （01字典树删除操作）

Problem Description
John is a manager of a CPU chip factory, the factory produces lots of chips everyday. To manage large amounts of products, every processor has a serial number. More specifically, the factory produces n chips today, the i-th chip produced this day has a serial number si.

At the end of the day, he packages all the chips produced this day, and send it to wholesalers. More specially, he writes a checksum number on the package, this checksum is defined as below:

which i,j,k are three different integers between 1 and n. And ⊕ is symbol of bitwise XOR.

Can you help John calculate the checksum number of today?

Input
The first line of input contains an integer T indicating the total number of test cases.

The first line of each test case is an integer n, indicating the number of chips produced today. The next line has n integers s1,s2,..,sn, separated with single space, indicating serial number of each chip.

1≤T≤1000
3≤n≤1000
0≤si≤109
There are at most 10 testcases with n>100

Output
For each test case, please output an integer indicating the checksum number in a line.

Sample Input
2
3
1 2 3
3
100 200 300

Sample Output
6
400

大致题意：给你n个数，让你从中选择三个不同的数，将其中两个数相加后异或上第三个数，使得结果最大，问最大结果为多少。

思路：先将这n个数建立成一颗01字典树，然后O（n^2）的去枚举两个不同的数，将其从字典树中删除，然后再查询此时异或最大值，然后再向字典树中加入这两个数。

代码如下

1.数组

#include<iostream>#include<cstdio>#include<algorithm>  #include<cstring>using namespace std;#define LL long long #define ULL unsigned long long const int MAXN=32*1005;struct Trie{    int ch[2];    int size;}T[MAXN];int root=1,tot=1;void add(int x){    int p=root;    T[p].size++;    for(int i=31;i>=0;i--)    {        int k=(x>>i)&1;        if(!T[p].ch[k]) T[p].ch[k]=++tot;        p=T[p].ch[k];        T[p].size++;        }}void Delete(int x){    int p=root;    T[p].size--;    for(int i=31;i>=0;i--)    {        int k=(x>>i)&1;        p=T[p].ch[k];        T[p].size--;    }}int query(int x){    int ans=0;    int p=root;    for(int i=31;i>=0;i--)    {        int k=(x>>i)&1;        if(T[p].ch[k^1]&&T[T[p].ch[k^1]].size)            ans^=(1<<i),p=T[p].ch[k^1];             else             p=T[p].ch[k];    }    return ans;}void init(){    for(int i=1;i<=tot;i++)    T[i].ch[0]=T[i].ch[1]=T[i].size=0;    tot=1;}int a[1005];int main() {    int T;    scanf("%d",&T);    int n,num;    for(int cas=1;cas<=T;cas++)    {        init();        int ans=-1;        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            add(a[i]);        }        for(int i=1;i<=n;i++)        {            Delete(a[i]);            for(int j=i+1;j<=n;j++)            {                Delete(a[j]);                ans=max(ans,query(a[i]+a[j]));                add(a[j]);            }            add(a[i]);        }        printf("%d\n",ans);    }    return 0;}

2.指针

#include<iostream>#include<cstdio>#include<algorithm>  #include<cstring>using namespace std;#define LL long long #define ULL unsigned long long struct node{    int size;    node *Next[2];};void add(node *head,int num){    node *p=head;    p->size++;    for(int i=31;i>=0;i--)    {        int k=(num>>i)&1;        if(p->Next[k]==NULL)        {            node *q=new node();            p->Next[k]=q;        }        p=p->Next[k];        p->size++;    }}void Delete(node *head,int num){    node *p=head;    p->size--;    for(int i=31;i>=0;i--)    {        int k=(num>>i)&1;        p=p->Next[k];        p->size--;    }}int query(node *head,int num){    node *p=head;    int x=0;    for(int i=31;i>=0;i--)    {        int k=(num>>i)&1;        if(p->Next[k^1]&&p->Next[k^1]->size)            p=p->Next[k^1],x^=(1<<i);        else         p=p->Next[k];    }    return x;}int a[1005];int main() {    int T;    scanf("%d",&T);    int n,num;    for(int cas=1;cas<=T;cas++)    {        node *head=new node();        int ans=-1;        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            add(head,a[i]);        }        for(int i=1;i<=n;i++)        {            Delete(head,a[i]);            for(int j=i+1;j<=n;j++)            {                Delete(head,a[j]);                ans=max(ans,query(head,a[i]+a[j]));                add(head,a[j]);            }            add(head,a[i]);        }        printf("%d\n",ans);    }    return 0;}