hdu 5536 Chip Factory 字典树

Chip Factory

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 427    Accepted Submission(s): 230

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:

maxi,j,k(si+sj)⊕sk

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

231 2 33100 200 300

 

Sample Output

6400

 

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

 

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解题报告：

1≤T≤1000
3≤n≤1000
0≤si≤109

There are at most 10 testcases with n>100

题意：输入n个数，求最大的(a+b)^c。（a、b、c要互不相同。）

建立字典树，把一个int型的数存放到字典树，里面每个结点有两条路径，nex[0]和nex[1]，代表这个结点的下一个节点是0还是1，并且有一个cnt表示有多少个数 在该位置（位运算，该结点表示的位置）是0或1（和该结点表示的数字 相同）。

字典树深度较小的点，表示的位数越高。

由于n最多有1000个，字典树的深度是32，那么数组大小超过1000*32即可。

for两重循环，表示a和b，如果a和b序号相同，则跳过。

然后在字典树里面，把a和b经过的每个结点的cnt--。（就是在字典树里拆掉a和b。）

然后遍历字典树，找到最优解:如果在树的某一层上，(a+b)在该位是1，那么找这一层有没有0的，若是0，则找1（贪心）。

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<climits>#include<queue>#include<vector>#include<map>#include<sstream>#include<set>#include<stack>#include<cctype>#include<utility>#pragma comment(linker, "/STACK:102400000,102400000")#define PI 3.1415926535897932384626#define eps 1e-10#define sqr(x) ((x)*(x))#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)#define  lson   num<<1,le,mid#define rson    num<<1|1,mid+1,ri#define MID   int mid=(le+ri)>>1#define zero(x)((x>0? x:-x)<1e-15)#define mk    make_pair#define _f     first#define _s     secondusing namespace std;//const int INF=    ;typedef long long ll;//const ll inf =1000000000000000;//1e15;//ifstream fin("input.txt");//ofstream fout("output.txt");//fin.close();//fout.close();//freopen("a.in","r",stdin);//freopen("a.out","w",stdout);const int INF =0x3f3f3f3f;const int maxn=1010    ;//const int maxm=    ;int n,root,cnt,a[maxn];struct Node{    int num;    int nex[2];     Node()    {        num=nex[1]=nex[0]=0;//严重错误:num=nex[1]=nex[2]=0;    };} node[maxn*33];int newNode(){    node[++cnt]= Node();    return cnt;}void build(int x){    int s=root, y;    for(int i=30;i>=0;i--)    {        if(x&(1<<i))  y=1;        else y=0;        /*这个是极端错误的写法！！！: int y= (x&(1<<i) );*/        Node & now=node[s];        if(now.nex[y]==0)  now.nex[y]=newNode();        s=now.nex[y];        node[s].num++;    }}void delet(int x){    int s=root,y;    for(int i=30;i>=0;i--)    {        if(x&(1<<i))  y=1;        else y=0;        Node & now=node[s];        s=now.nex[y];        node[s].num--;    }}int query(int x){    int s=root,y;    for(int i=30;i>=0;i--)    {        if(x&(1<<i))  y=1;        else y=0;        Node &now=node[s];        int ne=y;        //下面的if else 原来出错严重，在于ne的取值        if(y)        {            if(now.nex[0]&&  node[now.nex[0]].num)  ne=0  ;            else    x^=  ( 1<<i) ;        }        else        {            if(now.nex[1]&& node[now.nex[1]].num)  x^=  (1<<i)  ,ne=1 ;            else     ;        }        s=now.nex[ne];    }    return x;}int main(){    int T;    scanf("%d",&T);    while(T--)    {       scanf("%d",&n);       cnt=0;       root=newNode();       for(int i=1;i<=n;i++)  scanf("%d",&a[i]),build(a[i]);       int maxi=0;       for(int i=1;i<=n;i++)       {           delet(a[i]);           for(int j=1;j<=n;j++)           {               if(j==i)  continue;               delet(a[j]);               maxi=max(maxi,query(a[i]+a[j]) );               build(a[j]);           }           build(a[i]);       }       printf("%d\n",maxi);       for(int i=1;i<=cnt;i++)       {           node[i].num=node[i].nex[0]=node[i].nex[1]=0;       }    }    return 0;}