HDU5536 Chip Factory 【字典树】

Chip FactoryTime Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 2620    Accepted Submission(s): 1161Problem DescriptionJohn 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)⊕skwhich 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?InputThe 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≤10003≤n≤10000≤si≤109There are at most 10 testcases with n>100OutputFor each test case, please output an integer indicating the checksum number in a line.Sample Input231 2 33100 200 300Sample Output6400Source2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

这题数据弱，可以O(n3)卡过去

正经解法：
将每个数转化为二进制值，插入到字典树中
枚举val=～(a[i]+a[j]),并在字典树中删除对应的a[i],a[j]
val的第i位=x，如果字典树中存在第i位==x的 −−>ans+=(1<<i);
否则，查找字典树第i位==x^1

#include<iostream>#include<cstdlib>#include<cstdio>#include<string>#include<vector>#include<deque>#include<queue>#include<algorithm>#include<set>#include<map>#include<stack>#include<ctime>#include<string.h>#include<math.h>#include<list>using namespace std;#define ll long long#define pii pair<int,int>const int inf = 1e9 + 7;const int N = 1e3+5;inline int read(){    int x;    char ch;    while(!isdigit(ch=getchar()));    x=ch-'0';    while(isdigit(ch=getchar())){        x=x*10+ch-'0';    }    return x;}int a[N];const int CHARSET=2,BASE='0',MAX_NODE=32*1000;struct Trie{    int tot,root,child[MAX_NODE][CHARSET];    int flag[MAX_NODE];//flag[i]>0 存在 flag[i]=0:被删了    Trie(){        memset(child[1],0,sizeof(child[1]));        flag[1]=0;        root=tot=1;    }    void insert(int val){        int*cur=&root;        for(int i=30;i>=0;--i){            int x=(val&(1<<i))>0;            cur=&child[*cur][x];            if(*cur==0){                *cur=++tot;                memset(child[tot],0,sizeof(child[tot]));                flag[tot]=0;            }            ++flag[*cur];        }    }    void erase(int val){        int*cur=&root;        for(int i=30;i>=0;--i){            int x=(val&(1<<i))>0;            cur=&child[*cur][x];            --flag[*cur];        }    }    int query(int val){        int*cur=&root;        int ans=0;        for(int i=30;i>=0;--i){            int x=(val&(1<<i))>0;            int*tmp=&child[*cur][x];            if(*tmp!=0&&flag[*tmp]>0){                ans+=(1<<i);                cur=tmp;            }            else{                cur=&child[*cur][x^1];            }        }        return ans;    }    void clear(){        memset(child[1],0,sizeof(child[1]));        flag[1]=0;        root=tot=1;    }}trie;int slove(int n){    int ans=0;    for(int i=0;i<n;++i){        trie.erase(a[i]);        for(int j=i+1;j<n;++j){            trie.erase(a[j]);            ans=max(ans,trie.query(~(a[i]+a[j])));            trie.insert(a[j]);        }        trie.insert(a[i]);    }    return ans;}int main(){    //freopen("/home/lu/Documents/r.txt","r",stdin);    //freopen("/home/lu/Documents/w.txt","w",stdout);    int T=read();    while(T--){        int n=read();        trie.clear();        for(int i=0;i<n;++i){            a[i]=read();            trie.insert(a[i]);        }        printf("%d\n",slove(n));    }    return 0;}