DNA Prefix 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.

还是套用字典树的模板，稍微不同的地方是这题求前缀长度×字符串个数的最大值

那么再开一个数组pr来储存前缀长度

vis数组表示有多少个字符串共用此前缀

最大值ans = pr[i[*vis[i]

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;
#define N 1000010   //数组开大一点，否则runtime error
int tire[N][4];
int vis[N];
char s[60];
int pr[N];
int k = 1, ans;

int change(char s){
    if(s == 'A') return 0;
    else if(s == 'G') return 1;
    else if(s == 'C') return 2;
    else return 3;
}

void build(){
    int n = strlen(s);
    int p = 0;
    for(int i = 0; i<n; i++){
    int c = change(s[i]);
    if(!tire[p][c]){
        tire[p][c] = k;
        pr[k] = i + 1;
        k++;
    }
    p = tire[p][c];
    vis[p]++;
    }
}

int main()
{
    int t, n, te;
    scanf("%d", &t);
    te = t;
    while(t--){
        ans = 0;
        k = 1;
        memset(vis, 0, sizeof(vis));
        memset(tire, 0, sizeof(tire));
        scanf("%d", &n);
        while(n--){
            scanf("%s", s);
            build();
        }
    for(int i = 1; i<k; i++){
        ans = max(ans, pr[i]*vis[i]);
    }
    printf("Case %d: %d\n", te - t, ans);
    }
    return 0;
}