HDU5536 Chip Factory 【字典树】

Chip Factory Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 2620 Accepted Submission(s): 1161 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 Source 2015ACM/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; }