HDU6059 Kanade's trio（异或字典树）

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Kanade's trio

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1067 Accepted Submission(s): 392

Problem Description

Give you an array

A[1..n]，you need to calculate how many tuples

(i,j,k) satisfy that

(i<j<k) and

((A[i] xor A[j])<(A[j] xor A[k]))

There are T test cases.

1≤T≤20

1≤∑n≤5∗105

0≤A[i]<230

Input

There is only one integer T on first line.

For each test case , the first line consists of one integer

n ,and the second line consists of

n integers which means the array

A[1..n]

Output

For each test case , output an integer , which means the answer.

Sample Input

151 2 3 4 5

Sample Output

Source

2017 Multi-University Training Contest - Team 3

Recommend

liuyiding

http://blog.csdn.net/DorMOUSENone/article/details/76570172?locationNum=2&fps=1 看的是这位大佬的博客学的。

题意：给一个数列，找出多少个这样的有序三元组<i,j,k>满足a[i] xor a[j] < a[j] xor a[k]

#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int N = 500000 + 10;const int Node_max = N * 31;int T, n, num[30], a[N], cnt[31][2];long long ext = 0, ans;struct Node{    int nxt[2];    int cnt, ext;} Trie[Node_max];int Tsize;void calc(int tmp, long long c) {    //当前节点往下走有多少个字符串，从中间选两个    ans += Trie[tmp].cnt * 1ll * (Trie[tmp].cnt - 1) / 2;    //c代表的是这一位上，和初始字符不用的字符个数    ext += (c-Trie[tmp].cnt) * 1ll * Trie[tmp].cnt - Trie[tmp].ext;}void Trie_insert(int idx){    int tmp = 0;    for(int i=0;i<30;i++){                if(!Trie[tmp].nxt[ num[i] ]) {            Trie[tmp].nxt[ num[i] ] = ++Tsize;        }        //如果另一条路能走        if(Trie[tmp].nxt[ 1-num[i] ]) {            calc(Trie[tmp].nxt[ 1-num[i] ], cnt[i][ 1-num[i] ]);        }        tmp = Trie[tmp].nxt[ num[i] ];        Trie[tmp].cnt++;        //        Trie[tmp].ext += cnt[i][num[i]] - Trie[tmp].cnt;    }    return;}int main(){    scanf("%d", &T);    while(T-- && scanf("%d", &n)!=EOF)    {        memset(Trie, 0, Tsize * 16 + 16);        memset(cnt, 0, sizeof(cnt));        Tsize = 0;        ans = 0;        ext = 0;        for(int i=1, tmp;i<=n;i++)        {            scanf("%d", &a[i]);            tmp = a[i];            for(int j=29;j>=0;j--) {                num[j] = tmp%2;                cnt[j][tmp%2]++;//记录每一位为0或者为1的有多少个                tmp /= 2;            }            Trie_insert(i);        }        printf("%lld\n", ans + ext);    }}

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