LightOJ 1224

题目链接：https://cn.vjudge.net/problem/LightOJ-1224

Given a set of $n$ DNA samples, where each sample is a string containing characters from {A, C, G, T}, we are trying to find a subset of samples in the set, where the length of the longest common prefix multiplied by the number of samples in that subset is maximum.

To be specific, let the samples be:

ACGT

ACGTGCGT

ACCGTGC

ACGCCGT

If we take the subset {ACGT} then the result is 4 (4 * 1), if we take {ACGT, ACGTGCGT, ACGCCGT} then the result is 3 * 3 = 9 (since ACG is the common prefix), if we take {ACGT, ACGTGCGT, ACCGTGC, ACGCCGT} then the result is 2 * 4 = 8.

Now your task is to report the maximum result we can get from the samples.

Input
Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 50000) denoting the number of DNA samples. Each of the next n lines contains a non empty string whose length is not greater than 50. And the strings contain characters from {A, C, G, T}.

Output
For each case, print the case number and the maximum result that can be obtained.

Sample Input
3

4

ACGT

ACGTGCGT

ACCGTGC

ACGCCGT

3

CGCGCGCGCGCGCCCCGCCCGCGC

CGCGCGCGCGCGCCCCGCCCGCAC

CGCGCGCGCGCGCCCCGCCCGCTC

2

CGCGCCGCGCGCGCGCGCGC

GGCGCCGCGCGCGCGCGCTC

Sample Output
Case 1: 9

Case 2: 66

Case 3: 20

Note
Dataset is huge. Use faster I/O methods.

题意：

给出 $n$ 个字符串（只包含四个字符：$A,C,G,T$），让你从中挑选出 $k$ 个字符串，设他们的最长公共前缀的长度为 $L$。求最大的 $k cdot L$。

题解：

字典树的每个节点开一个 $pre$ 记录该节点对应的前缀是多少个字符串的前缀。

然后DFS遍历Trie的每个节点，维护节点深度乘节点 $pre$ 值的最大值即可。时间复杂度等于字典树的空间复杂度。

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=5e4+5;
int n;
char s[53];

namespace Trie
{
    const int SIZE=maxn*50;
    int sz;
    struct TrieNode{
        int pre,ed;
        int nxt[4];
    }trie[SIZE];
    void init()
    {
        sz=1;
        memset(trie,0,sizeof(trie));
    }
    inline int idx(const char& c)
    {
        if(c=='A') return 0;
        if(c=='C') return 1;
        if(c=='G') return 2;
        if(c=='T') return 3;
    }
    void insert(const string& s)
    {
        int p=1;
        for(int i=0;i<s.size();i++)
        {
            int ch=idx(s[i]);
            if(!trie[p].nxt[ch]) trie[p].nxt[ch]=++sz;
            p=trie[p].nxt[ch];
            trie[p].pre++;
        }
        trie[p].ed++;
    }

    int ans;
    void dfs(int d,int now)
    {
        if(!now) return;
        ans=max(ans,d*trie[now].pre);
        for(int i=0;i<4;i++) dfs(d+1,trie[now].nxt[i]);
    }
};

int main()
{
    int T;
    cin>>T;
    for(int kase=1;kase<=T;kase++)
    {
        scanf("%d",&n);
        Trie::init();
        for(int i=1;i<=n;i++)
        {
            scanf("%s",s);
            Trie::insert(s);
        }
        Trie::ans=0;
        Trie::dfs(0,1);
        printf("Case %d: %d
",kase,Trie::ans);
    }
}