标签：数据结构-字典树

Prefix goodness of a set string is length of longest common prefix*number of strings in the set. For example the prefix goodness of the set {000,001,0011} is 6.You are given a set of binary strings. Find the maximum prefix goodness among all possible subsets of these binary strings.

Input
First line of the input contains T(≤20) the number of test cases. Each of the test cases start with n(≤50000) the number of strings. Each of the next n lines contains a string containing only 0 and 1. Maximum length of each of these string is 200.

Output
For each test case output the maximum prefix goodness among all possible subsets of n binary strings.

Sample Input                             Output for Sample Input

字典树，每条边记录一下深度。

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <string>
using namespace std;

struct edge{
    int u , v , letter , sum , dep;
    edge(int a = 0 , int b = 0 , int c = 0 , int d = 0 , int f = 0){
        u = a , v = b , letter = c , sum = d , dep = f;
    }
};
vector<edge> e;
vector<int> head , next;
int ans;

void add(int  u , int v , int letter , int sum , int dep){
    e.push_back(edge(u , v , letter , sum , dep));
    next.push_back(head[u]);
    head[u] = next.size()-1;
    head.push_back(-1);
}

void addLibrary(string s){
    int u = 0 , index = 0;
    while(index < s.length()){
        int Next = head[u];
        while(Next != -1){
            if(e[Next].letter == s[index]-'0') break;
            Next = next[Next];
        }
        if(Next == -1) break;
        e[Next].sum++;
        index++;
        u = e[Next].v;
    }
    while(index < s.length()){
        add(u , head.size() , s[index]-'0' , 1 , index+1);
        index++;
        u = head.size()-1;
    }
}
void initial(){
    e.clear();
    next.clear();
    head.clear();
    head.push_back(-1);
    ans = 0;
}

void readcase(){
    string s;
    int n;
    scanf("%d" , &n);
    getchar();
    while(n--){
        cin >> s;
        //cout <<"+"<<s<<endl;
        addLibrary(s);
    }
    for(int i = 0; i < e.size(); i++){
        if(e[i].sum*e[i].dep > ans) ans = e[i].sum*e[i].dep;
    }
    printf("%d\n" , ans);
}

int main(){
    int t;
    scanf("%d" , &t);
    while(t--){
        initial();
        readcase();
    }
    return 0;
}

uva 11488 - Hyper Prefix Sets（字典树）,布布扣,bubuko.com

uva 11488 - Hyper Prefix Sets（字典树）

标签：数据结构-字典树

原文地址：http://blog.csdn.net/u011836218/article/details/32729371