HDU 5536 (ACM 2015 长春) Chip Factory [Trie树]

题意:给出N个数,选出三个下标不同的数,令(Ai+Aj)^Ak最大。

范围:N<=1000

解法:首先想到两数异或和最大可以贪心+Trie树搜索解决,此时只需要将所有数化为01串加入Trie树,并且枚举i,j,在trie树中贪心得到一个异或或最大的即可(要先把i,j从Trie树中删除,计算完后再insert)

代码:

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<iostream>
#include<stdlib.h>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<bitset>
#pragma comment(linker, "/STACK:1024000000,1024000000")
template <class T>
bool scanff(T &ret){ //Faster Input
    char c; int sgn; T bit=0.1;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    if(c==' '||c=='\n'){ ret*=sgn; return 1; }
    while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
    ret*=sgn;
    return 1;
}
#define inf 1073741823
#define llinf 4611686018427387903LL
#define PI acos(-1.0)
#define lth (th<<1)
#define rth (th<<1|1)
#define rep(i,a,b) for(int i=int(a);i<=int(b);i++)
#define drep(i,a,b) for(int i=int(a);i>=int(b);i--)
#define gson(i,root) for(int i=ptx[root];~i;i=ed[i].next)
#define tdata int testnum;scanff(testnum);for(int cas=1;cas<=testnum;cas++)
#define mem(x,val) memset(x,val,sizeof(x))
#define mkp(a,b) make_pair(a,b)
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define pb(x) push_back(x)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

#define NN 100100

int n,a[NN],tot;
struct node{
    int cot;
    int a[2];
    void init(){
        a[0]=a[1]=0;
        cot=0;
    }
}t[110110];
void init(){
    tot=0;
    t[0].init();
}
int b[55];
void insert(int x){
    int y=1;
    rep(i,1,31){
        if(x&y)b[i]=1;
        else b[i]=0;
        y<<=1;
    }
    reverse(b+1,b+32);
    x=0;
    t[x].cot++;
    rep(i,1,31){
        if(t[x].a[b[i]])x=t[x].a[b[i]];
        else{
            t[++tot].init();
            t[x].a[b[i]]=tot;
            x=tot;
        }
        t[x].cot++;
    }
}
void dele(int x){
    int y=1;
    rep(i,1,31){
        if(x&y)b[i]=1;
        else b[i]=0;
        y<<=1;
    }
    reverse(b+1,b+32);
    x=0;
    t[x].cot--;
    rep(i,1,31){
        x=t[x].a[b[i]];
        t[x].cot--;
    }
}
int find(int x){
    int y=1;
    rep(i,1,31){
        if(x&y)b[i]=1;
        else b[i]=0;
        y<<=1;
    }
    reverse(b+1,b+32);

    x=0;
    int ans=0;
    rep(i,1,31){
        if(t[x].cot>0){
            int l=t[x].a[b[i]^1];
            int r=t[x].a[b[i]];
            if(l&&t[l].cot>0){
                x=l;
                ans=(ans<<1)|(b[i]^1);
            }
            else if(r&&t[r].cot>0){
                x=r;
                ans=(ans<<1)|(b[i]);
            }

        }

    }

    return ans;
}
int main()
{
    tdata{
        init();
        scanff(n);
        rep(i,1,n)scanff(a[i]),insert(a[i]);
        int ans=0;
        rep(i,1,n){
            dele(a[i]);
            rep(j,i+1,n){
                dele(a[j]);
                int x=a[i]+a[j];
                int y=find(x);
                ans=max(ans,x^y);
                insert(a[j]);
            }
            insert(a[i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}

    原文作者:Trie树
    原文地址: https://blog.csdn.net/GrassTreeFlower/article/details/49561947
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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