4103: [Thu Summer Camp 2015]异或运算

4103: [Thu Summer Camp 2015]异或运算

Time Limit: 20 Sec Memory Limit: 512 MB
Submit: 566 Solved: 302
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Description

给定长度为n的数列X={x1,x2,…,xn}和长度为m的数列Y={y1,y2,…,ym}，令矩阵A中第i行第j列的值Aij=xi xor yj，每次询问给定矩形区域i∈[u,d],j∈[l,r]，找出第k大的Aij。

Input

第一行包含两个正整数n,m，分别表示两个数列的长度

第二行包含n个非负整数xi
第三行包含m个非负整数yj
第四行包含一个正整数p，表示询问次数
随后p行，每行均包含5个正整数，用来描述一次询问，每行包含五个正整数u,d,l,r,k，含义如题意所述。
Output

共p行，每行包含一个非负整数，表示此次询问的答案。

Sample Input

3 3

1 2 4

7 6 5

3

1 2 1 2 2

1 2 1 3 4

2 3 2 3 4
Sample Output

6

5

1
HINT

对于100%的数据，0<=Xi,Yj<2^31,

1<=u<=d<=n<=1000,

1<=l<=r<=m<=300000,

1<=k<=(d-u+1)*(r-l+1),

1<=p<=500
Source

鸣谢佚名上传

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注意到p，n都很小
每次询问，考虑从高到低逐位确定
于是就得解决，每次给一个前缀，询问有这个前缀的数字的数量
对于数列Y，把它建成一棵可持久化trie
每次需要确定前缀的时候，对于X数列的每个数字，都单独拿出来考虑
就在trie树上继续往下走一步然后做差就行了
复杂度O(pnlog(maxvalue))

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 32;const int maxn = 1005;const int maxm = 3E5 + 3;int n,m,q,cnt,mi[N],A[maxn],o1[maxn],o2[maxn],rt[maxm],siz[maxm*N],ch[maxm*N][2];inline int getint(){    char ch = getchar(); int ret = 0;    while (ch < '0' || '9' < ch) ch = getchar();    while ('0' <= ch && ch <= '9')        ret = ret * 10 + ch - '0',ch = getchar();    return ret;}inline void Solve(){    int l1,r1,l2,r2,k,Ans = 0;    l1 = getint(); r1 = getint();    l2 = getint(); r2 = getint(); k = getint();    k = (r1 - l1 + 1) * (r2 - l2 + 1) - k + 1;    for (int i = l1; i <= r1; i++)        o1[i] = rt[l2 - 1],o2[i] = rt[r2];    for (int i = 30; i >= 0; i--)    {        int sum = 0;        for (int j = l1; j <= r1; j++)        {            int tmp = (A[j] & mi[i]) ? 1 : 0;            sum += siz[ch[o2[j]][tmp]] - siz[ch[o1[j]][tmp]];        }        if (sum >= k)        {            for (int j = l1; j <= r1; j++)            {                int tmp = (A[j] & mi[i]) ? 1 : 0;                o2[j] = ch[o2[j]][tmp]; o1[j] = ch[o1[j]][tmp];            }        }        else        {            Ans |= mi[i]; k -= sum;            for (int j = l1; j <= r1; j++)            {                int tmp = (A[j] & mi[i]) ? 0 : 1;                o2[j] = ch[o2[j]][tmp]; o1[j] = ch[o1[j]][tmp];            }        }    }    printf("%d\n",Ans);}int main(){    #ifdef DMC        freopen("DMC.txt","r",stdin);    #endif    n = getint(); m = getint(); mi[0] = 1;    for (int i = 1; i < 31; i++) mi[i] = mi[i - 1] << 1;    for (int i = 1; i <= n; i++) A[i] = getint();    for (int i = 1; i <= m; i++)    {        int o1 = rt[i - 1],o2 = rt[i] = ++cnt,x = getint();        for (int j = 30; j >= 0; j--)        {            int Nex = (x & mi[j]) ? 1 : 0;            ch[o2][Nex^1] = ch[o1][Nex^1];            ch[o2][Nex] = ++cnt; o1 = ch[o1][Nex];            o2 = ch[o2][Nex]; siz[o2] = siz[o1] + 1;        }    }    q = getint(); while (q--) Solve();    return 0;}