Uva 11488 Hyper Prefix Sets 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
4
4
0000
0001
10101
010
2
01010010101010101010
11010010101010101010
3
010101010101000010001010
010101010101000010001000
010101010101000010001010
5
01010101010100001010010010100101
01010101010100001010011010101010
00001010101010110101
0001010101011010101
00010101010101001
Sample Output
6
20
66

44

题意：给出一个字符串的集合，找出一个子集，使得子集中所有的字符串的公共前缀的长度之和最长

数据结构：字典树

设计：每个节点存储这个节点被插入的次数cnt，建好树之后，进行一次深度优先遍历，对于每一个i节点，用它的深度乘以cnt，得到的值就是以这个节点为结尾的公共前缀长度之和。找出最大的值，别忘了回溯的时候释放内存，不然内存会boom掉

#include <iostream>#include <stdio.h>#define MAX 210using namespace std;struct trie{    int cnt;    trie *next[2];};trie *root=new trie;int n,res;char str[MAX];void insert_trie(){    trie *p,*newnode;    p=root;    for(int i=0; str[i]!='\0'; i++)    {        if(p->next[str[i]-'0']==NULL)        {            newnode=new trie;            newnode->cnt=1;            newnode->next[0]=newnode->next[1]=NULL;            p->next[str[i]-'0']=newnode;            p=newnode;        }        else        {            p->next[str[i]-'0']->cnt++;            p=p->next[str[i]-'0'];        }    }}void dfs(trie *p,int cur){    for(int i=0; i<2; i++)    {        if(p->next[i]!=NULL)        {            int tmp=cur*p->next[i]->cnt;            if(res<tmp)                res=tmp;dfs(p->next[i],cur+1);delete(p->next[i]);p->next[i]=NULL;        }    }}int main(){    int t,i;    scanf("%d",&t);    while(t--)    {        res=0;        for(i=0;i<2;i++)root->next[i]=NULL;        scanf("%d",&n);        for(i=0; i<n; i++){ scanf("%s",str); insert_trie();}//printf("finish insert\n");dfs(root,1);printf("%d\n",res);    }    return 0;}