hdu5536 Chip Factory xor,字典树
hdu5536 Chip Factory xor,字典树
Chip Factory
Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 280 Accepted Submission(s): 158
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?
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
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
看到异或就应该想到字典树。。。
注意用siz来处理删除,而不要直接删除结点。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<bits/stdc++.h> #define REP(i,a,b) for(int i=a;i<=b;i++) #define MS0(a) memset(a,0,sizeof(a)) using namespace std; typedef long long ll; const int maxn=1000100; const int INF=(1<<29); const double EPS=0.000000001; const double Pi=acos(-1.0); struct Trie { int ch[2]; int siz; void debug() { printf("lch=%2d rch=%2d siz=%2d\n",ch[1],ch[0],siz); } };Trie tr[maxn];int p;int rt; int n,a[maxn]; int newnode() { tr[++p].siz=0; tr[p].ch[0]=tr[p].ch[1]=-1; return p; } void Init() { rt=0; tr[rt].ch[0]=tr[rt].ch[1]=-1; tr[rt].siz=0; p=0; } void Insert(int t) { int k=rt; tr[k].siz++; int c; for(int i=30;i>=0;i--){ if(t&(1<<i)) c=1; else c=0; if(tr[k].ch[c]==-1) tr[k].ch[c]=newnode(); k=tr[k].ch[c]; tr[k].siz++; } } void Delete(int t) { int k=rt; tr[k].siz--; int c; for(int i=30;i>=0;i--){ if(t&(1<<i)) c=1; else c=0; k=tr[k].ch[c]; tr[k].siz--; } } int query(int t) { int k=rt; int res=0; for(int i=30;i>=0;i--){ if(t&(1<<i)){ if(tr[k].ch[0]==-1||tr[tr[k].ch[0]].siz==0){ k=tr[k].ch[1]; } else{ k=tr[k].ch[0]; res|=(1<<i); } } else{ if(tr[k].ch[1]==-1||tr[tr[k].ch[1]].siz==0){ k=tr[k].ch[0]; } else{ k=tr[k].ch[1]; res|=(1<<i); } } } return res; } int main() { freopen("in.txt","r",stdin); int T;cin>>T; while(T--){ scanf("%d",&n); Init(); REP(i,1,n) scanf("%d",&a[i]),Insert(a[i]); int ans=0; REP(i,1,n){ REP(j,i+1,n){ Delete(a[i]);Delete(a[j]); int tmp=query(a[i]+a[j]); ans=max(ans,tmp); Insert(a[i]);Insert(a[j]); } } cout<<ans<<endl; } return 0; }
没有AC不了的题,只有不努力的ACMER!