题目链接
题解
一眼看过去是二维结构，实则未然需要树套树之类的数据结构
区域异或和，就一定是可持久化(trie)树
观察数据，(m)非常大，而(n)和(p)比较小，甚至可以每次询问都枚举(x_i)
所以我们可以考虑对(y_i)建(trie)，每次询问取出对应区间的(x_i)在对应区间的(trie)树中跑
多点询问和单点询问时类似的，只不过它们会分开走
我们只需每次记录每个(x_i)所在的节点
对于每一层，统计一下能异或出多少(1)，如果(le k)，每个(x_i)往能异或出(1)的方向走，否则走另一边，并令(k)减去这些数
复杂度(O(31m + 31qn))
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 300005,B = 30,maxm = 10000005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
int bin[50],n,m;
int a[1005],b[maxn],rt[maxn],cnt;
int ch[maxm][2],sum[maxm];
int ins(int x,int v){
int u,tmp;
u = tmp = ++cnt;
for (int i = B; ~i; i--){
ch[u][0] = ch[v][0];
ch[u][1] = ch[v][1];
sum[u] = sum[v] + 1;
int t = bin[i] & x; t >>= i;
v = ch[v][t];
u = ch[u][t] = ++cnt;
}
sum[u] = sum[v] + 1;
return tmp;
}
int atu[1005],atv[1005];
int query(int u,int v,int l,int r,int k){
int ans = 0;
for (int i = l; i <= r; i++) atu[i] = u,atv[i] = v;
for (int i = B; ~i; i--){
int cnt = 0;
for (int j = l; j <= r; j++){
int x = atu[j],y = atv[j],t = bin[i] & a[j]; t >>= i;
cnt += sum[ch[x][t ^ 1]] - sum[ch[y][t ^ 1]];
}
if (cnt >= k){
ans += bin[i];
for (int j = l; j <= r; j++){
int x = atu[j],y = atv[j],t = bin[i] & a[j]; t >>= i;
atu[j] = ch[x][t ^ 1];
atv[j] = ch[y][t ^ 1];
}
}
else {
k -= cnt;
for (int j = l; j <= r; j++){
int x = atu[j],y = atv[j],t = bin[i] & a[j]; t >>= i;
atu[j] = ch[x][t];
atv[j] = ch[y][t];
}
}
}
return ans;
}
int main(){
bin[0] = 1; for (int i = 1; i <= 30; i++) bin[i] = bin[i - 1] << 1;
n = read(); m = read();
for (int i = 1; i <= n; i++) a[i] = read();
for (int i = 1; i <= m; i++){
b[i] = read();
rt[i] = ins(b[i],rt[i - 1]);
}
int p = read(),u,d,l,r,k;
while (p--){
u = read(); d = read(); l = read(); r = read(); k = read();
printf("%d
",query(rt[r],rt[l - 1],u,d,k));
}
return 0;
}