BOJ 487 xor

时间限制 5000 ms 内存限制 65536 KB

题目描述

Given a sequence A1,A2,…,AN, you're required to answer the following queries: given three integers l,r and x, find out max{ai⊕x}, where l≤i≤r, and a⊕b indicates the bitwise XOR(xor) operation.

输入格式

An integer T(T≤10) will exist in the first line of input, indicating the number of test cases. Each test case begins with two integers Nand M(1≤N,M≤50000), which separately denotes the length of the sequence and the number of queries. The following line gives N integers from A1 to AN. The next M lines, each with three integers l,r,x(1≤l≤r≤N), describe all the queries. All the elements in sequence A and the queried x will be set in range [0,10000].

输出格式

Output the answer for each query, one per line.

输入样例

13 21 3 21 2 11 3 1

输出样例

23

题意：

给你一段长度为 N 的数列，再给你 M 个查询， 问在 L ~ R 区间内与 x 异或最大的值

解题思路：

首先，我们明确地知道求一段区间内异或值最大的用Trie树查询，就是以该位的非值作为最优解查询，如果没有最优解不存在，或者不满足条件，则选择次优解，即该位本身数值。

其次是区间查询，在这里采用离线查询的方式，将查询按照R的大小查询，能够保证每一次查询的都满足小于R区间。L区间则是在Trie的find中维护，Trie的val数组维护的每一个经过该点的最大角标，然后就按照最优解是否存在，最优解是否在L的右边查询。

因为个人写的是递归的find，所以需要注意最后一位的处理。（之前没有判断最后一位的最优性....但是跑随机数都是对的

#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#define eps 1e-9#define INF 0x3f3f3f3fusing namespace std;typedef long long ll;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;template<class T>inline bool read(T &n){    T x = 0, tmp = 1; char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}template <class T>inline void write(T n){    if(n < 0)    {        putchar('-');        n = -n;    }    int len = 0,data[20];    while(n)    {        data[len++] = n%10;        n /= 10;    }    if(!len) data[len++] = 0;    while(len--) putchar(data[len]+48);}//-----------------------------------const int MAXN=500010;struct Node{    int l,r,x;    int idx,ans;}a[MAXN];int num[MAXN],ans_flag=0;int n,q;struct Trie{int ch[2*MAXN][2];int v[2*MAXN];int sz;Trie(){ sz=1; }int idx(int x){ return x%2; }void insert(int x,int po){int u=0;for(int i=13;i>=0;i--){int c=idx(x>>i);//cout<<(x>>i)<<endl;if(!ch[u][c]){v[sz]=po;ch[u][c]=sz++;}v[u]=po;//            cout<<" "<<u<<" "<<sz<<" "<<v[u]<<endl;u=ch[u][c];}v[u]=po;//cout<<x<<endl;}int find(int x,int l,int po,int i){    if(i==-1&&v[po]>=l)            return v[po];        int c=idx(x>>i);//cout<<po<<" "<<i<<" "<<c<<" "<<v[po]<<endl;system("pause");        if(ch[po][!c])        {            if(v[ch[po][!c]]>=l)                return find(x,l,ch[po][!c],i-1);            else                return find(x,l,ch[po][c],i-1);        }        return find(x,l,ch[po][c],i-1);}void clear(){    sz=1;memset(ch,0,sizeof(ch));memset(v,0,sizeof(v));}};Trie s;bool cmp(Node a,Node b){    return a.r<b.r;}bool cmp1(Node a,Node b){    return a.idx<b.idx;}int main(){    int T;//    freopen("data.txt","r",stdin);    read(T);    while(T--)    {        s.clear();        read(n),read(q);        for(int i=1;i<=n;i++)            read(num[i]);        for(int i=1;i<=q;i++)        {            read(a[i].l);read(a[i].r);read(a[i].x);            a[i].idx=i;        }        sort(a+1,a+q+1,cmp);        int nr=1;        for(int i=1;i<=q;i++)        {            ans_flag=0;            for(int j=nr;j<=a[i].r;j++)                s.insert(num[j],j);            nr=a[i].r+1;            a[i].ans=a[i].x^num[s.find(a[i].x,a[i].l,0,13)];//cout<<"  "<<a[i].ans<<endl;        }        sort(a+1,a+q+1,cmp1);        for(int i=1;i<=q;i++)            printf("%d\n",a[i].ans);    }    return 0;}