HDU 5536 Chip Factory(01字典树找最大异或值)——2015ACM/ICPC亚洲区长春站
2015-11-01 19:22
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Chip Factory
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
/*********************************************************************/
有一个长度为n(n≤1000)的序列s,对于不同的i,j,k,求
maxi,j,k(si+sj)⊕sk
其中⊕代表异或运算。
解题思路:据说现场赛的时候O(n3)是可以过的,然而现场赛重现的时候会送你一个可爱的TLE
此题的正规解法是建一棵字典树,就是Trie树
对s构一个Trie树,然后枚举i,j,每次删除si,sj再查询,再插入回来就可以了。
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std; const int MAXN=100010; int T_T,n,a[MAXN]; struct Trie { int ch[2],size; }T[MAXN]; int root=1,tot=1; void Insert(int x) { int o=root; T[o].size++; for(int k=30;k>=0;k--) { int c; if(x&(1<<k)) c=1; else c=0; if(!T[o].ch[c]) T[o].ch[c]=++tot; o=T[o].ch[c]; T[o].size++; } } void Delete(int x) { int o=root; T[o].size--; for(int k=30;k>=0;k--) { int c; if(x&(1<<k)) c=1; else c=0; o=T[o].ch[c]; T[o].size--; } } int Query(int x) { int o=root; for(int k=30;k>=0;k--) { int c; if(x&(1<<k)) c=1; else c=0; if(c==1) { if(T[o].ch[0]&&T[T[o].ch[0]].size) o=T[o].ch[0]; else o=T[o].ch[1],x^=(1<<k); } else { if(T[o].ch[1]&&T[T[o].ch[1]].size) o=T[o].ch[1],x^=(1<<k); else o=T[o].ch[0]; } } return x; } int main() { scanf("%d",&T_T); while(T_T--) { int ans=0; scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&a[i]); for(int i=1;i<=n;i++) Insert(a[i]); for(int i=1;i<=n;i++) { Delete(a[i]); for(int j=i+1;j<=n;j++) { Delete(a[j]); ans=max(ans,Query(a[i]+a[j])); Insert(a[j]); } Insert(a[i]); } printf("%d\n",ans); for(int i=1;i<=tot;i++) T[i].ch[0]=0,T[i].ch[1]=0,T[i].size=0; tot=1; } return 0; }菜鸟成长记
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