HDU-5536-Chip Factory【字典树】
Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
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
2
3
1 2 3
3
100 200 300
Sample Output
6
400
题目链接：HDU-5536
题目大意：n个数字的序列中，找出三个数字使得（a[i] + a[j]）^a[k]最大。
题目思路：现场赛的时候，一直超时没过。听学长说可以用n^3的方法暴力优化过，不过正确的解法是用01字典树。
把这n个数字保存下来建在一个01字典树上面。因为i,j,k三个数字不能重复，所以删去要用的i和j，再在里面找出能和num( a[i]+a[j])异或出的最大值。
如果num这一位上是1，就去树上找对应位的0，如果存在就由该点0往下继续找.如果不存在0就找1.（保证找出来的k值和num异或最大）
以下是代码：
#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <cstring>
using namespace std;
#define MAXN 4000005
struct Trie // 定义01Trie
{
int next[2];
int flag,number;
void ini() //初始化
{
next[0] = next[1] = 0;
flag = 0;
}
}pnode[MAXN];
int root = 0;
int tot = 0;
void Insert(int num)
{
int rt = root;
for (int i = 30; i >= 0; i--)
{
int ret;
if (num & (1 << i)) ret = 1; //得到第i位是1还是0
else ret = 0;
if (!pnode[rt].next[ret]) //如果没有这个节点
{
pnode[rt].next[ret] = ++tot; //新增加这个节点，并且把这个子节点标记为tot
pnode[tot].number = ret; //tot这个子节点的值为ret
}
rt = pnode[rt].next[ret]; // 获得子节点的编号
pnode[rt].flag++; //该节点所经过的flag++；
}
}
void Delet (int num) //删去该节点
{
int rt = root;
for (int i = 30; i >= 0; i--)
{
int ret;
if (num & (1 << i)) ret = 1;
else ret = 0;
rt = pnode[rt].next[ret];
pnode[rt].flag--; //该节点所经过的这条路上flag--；
}
}
int Query(int num)
{
int rt = root;
for (int i = 30; i >= 0; i--)
{
int ret;
if (num & (1 << i)) ret = 1;
else ret = 0;
if (ret == 1) //如果为1，子节点先搜0,0不存在再走1.
{
int next_node = pnode[rt].next[0];
if (pnode[next_node].flag > 0 && next_node) rt = pnode[rt].next[0]; //0存在
else
{
rt = pnode[rt].next[1];
num ^= (1 << i);
}
}
else //如果为0，子节点先搜1,1不存在再走0.
{
int next_node = pnode[rt].next[1];
if (pnode[next_node].flag > 0 && next_node) //1存在
{
rt = pnode[rt].next[1];
num ^= (1 << i);
}
else rt = pnode[rt].next[0];
}
}
return num;
}
int a[MAXN];
int main(){
int t;
scanf("%d",&t);
while(t--)
{
for(int i = 0;i < tot;i++) //初始化
{
pnode[i].ini();
}
int n,ans = 0;
scanf("%d",&n);
for (int i = 0; i < n; i++)
{
scanf("%d",&a[i]);
Insert(a[i]);
}
for (int i = 0; i < n; i++)
{
Delet(a[i]); //不存在相同的i,j,k所以先把i和j从字典树中删除
for (int j = i + 1; j < n; j++)
{
Delet(a[j]);
ans = max(Query(a[i] + a[j]),ans);
Insert(a[j]);
}
Insert(a[i]);
}
printf("%d\n",ans);
}
return 0;
}