Rikka with Stable Marriage
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 242 Accepted Submission(s): 145
Problem Description
People in love always feel humble. Sometimes, Rikka is worried about whether she deserves love from Yuta.
Stable marriage problem is an interesting theoretical model which has a strong connection with the real world. Given n men and n women, where each person has ranked all members of the opposite sex in order of preference. We use a permutation p of length n to represent a match that the ith man gets married with the pith woman. A match is stable if and only if there are no two people of opposite sexes who would both rather have each other than their current partners, i.e., ∀i≠j,¬(ri(pj,pi)∧rpj(i,j)) where ra(b,c) represents whether person a loves b more than c.
Rikka wants to resolve the confusion in her mind by considering a special case of the stable marriage problem. Rikka uses a feature integer to represents a person's character, and for two persons with feature integers equal to x and y, Rikka defines the suitable index of them as x⊕y, where ⊕ represents binary exclusive-or.
Given n men with feature integers a1,…,an and n women with feature integers b1,…,bn. For the ith man, he loves the jth woman more than the kth woman if and only if ai⊕bj>ai⊕bk; for the ith woman, she loves the jth man more than the kth man if and only if bi⊕aj>bi⊕ak.
Rikka wants to calculate the best stable match for this problem, i.e., let P be the set of all stable match, she wants to calculate maxp∈P(∑ni=1(ai⊕bpi)). Since n is quite large, this problem is too difficult for Rikka, could you please help her find the answer?
Input
The first line of the input contains a single integer T(1≤T≤50), the number of test cases.
For each test case, the fisrt line contains a sigle integer n(1≤n≤105).
The second line contains n integers a1,…,an(1≤ai≤109) which represents the feature number of each man.
The third line contains n integers b1,…,bn(1≤bi≤109) which represents the feature number of each woman.
The input guarantees that there are no more than 5 test cases with n>104, and for any i,j∈[1,n],i≠j, ai≠aj and bi≠bj.
Output
For each test case, output a single line with a single integer, the value of the best stable match. If there is no stable match, output −1.
Hint
In the first test case, one of the best matches is (2,1,4,3). Therefore the answer is (1⊕2)+(2⊕1)+(3⊕4)+(4⊕3)=20.
Sample Input
2
4
1 2 3 4
1 2 3 4
5
10 20 30 40 50
15 25 35 45 55
Sample Output
20
289
题意
有n个男生，n个女生（n<=100000），每个人有一个权值，每两个人的好感度为a[i]^b[j]，求将他们稳定婚姻匹配的最大权值。
题解
考场上只想到了O(n^2)。。。
正解是把两组人分别建两棵Trie树（其实也可以只建一棵），从根节点向下走
贪心地想一下，我们要让异或值最大的话，就应该使他们的路径尽量不同
所以优先使它们走相反方向，进行匹配，统计答案
直到不行时，再走相同方向，统计答案
（这个思路太巧妙了，代码又短）
代码：
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 100005
#define LOG 30
struct node{
int x,l,r,siz1,siz2;
void clear(){x=l=r=siz1=siz2=0;}
}a[N*LOG];
int tot,rt;
void pushup(int i)
{
a[i].siz1=a[a[i].l].siz1+a[a[i].r].siz1;
a[i].siz2=a[a[i].l].siz2+a[a[i].r].siz2;
}
void insert(int &i,int x,int dep,int flg)
{
if(!i)a[i=++tot].clear();
if(!dep){
a[i].x=x;
if(flg) a[i].siz1++;
else a[i].siz2++;
return;
}
if(x&(1<<(dep-1))) insert(a[i].r,x,dep-1,flg);
else insert(a[i].l,x,dep-1,flg);
pushup(i);
}
long long ans;
void match(int i,int j,int dep)
{
if(!a[i].siz1||!a[j].siz2) return;
if(!dep){
int tmp=min(a[i].siz1,a[j].siz2);
a[i].siz1-=tmp;a[j].siz2-=tmp;
ans+=1ll*tmp*(a[i].x^a[j].x);
return;
}
match(a[i].l,a[j].r,dep-1);
match(a[i].r,a[j].l,dep-1);
match(a[i].l,a[j].l,dep-1);
match(a[i].r,a[j].r,dep-1);
pushup(i);pushup(j);
}
int main()
{
int T,n,i,x;
scanf("%d",&T);
while(T--){
rt=tot=0;ans=0;
scanf("%d",&n);
for(i=1;i<=n;i++){
scanf("%d",&x);
insert(rt,x,30,1);
}
for(i=1;i<=n;i++){
scanf("%d",&x);
insert(rt,x,30,0);
}
match(rt,rt,30);
printf("%lld\n",ans);
}
}Trie树真好用