Chip Factory(01字典树)

Chip Factory

http://acm.hdu.edu.cn/showproblem.php?pid=5536

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:

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
 
用num标记该点是否被删除,剩下的就是01字典树的模板了
 
 #include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#define maxn 1005
using namespace std;
int tol;
long long val[maxn*];
int tree[maxn*][];
int num[maxn*][];
long long a[maxn]; void init(){
tol=;
tree[][]=tree[][]=;
memset(num,,sizeof(num));
} void add(long long x,int k){
int u=;
for(int i=;i>=;i--){
int v=(x>>i)&;
num[u][v]+=k;
if(!tree[u][v]){
tree[tol][]=tree[tol][]=;
val[tol]=-;
tree[u][v]=tol++;
}
u=tree[u][v];
}
val[u]=x;
} long long query(long long n){
int u=;
for(int i=;i>=;i--){
int v=(n>>i)&;
if(num[u][v^]) u=tree[u][v^];
else u=tree[u][v];
}
return val[u];
} int main(){
int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
init();
for(int i=;i<=n;i++){
scanf("%lld",&a[i]);
add(a[i],);
}
long long tmp;
long long ans=;
for(int i=;i<=n;i++){
add(a[i],-);
for(int j=i+;j<=n;j++){
add(a[j],-);
tmp=a[i]+a[j];
ans=max(ans,tmp^query(tmp));
add(a[j],);
}
add(a[i],);
}
printf("%lld\n",ans);
} }
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