数据结构 trie 树 入门练手 POJ 3630 Phone List

跟那个排序二叉树的那个静态树一样，不大习惯静态的，但是不静态会tle，94ms

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

typedef struct node
{
    int flag;
    int next[11];
}trie;
int leap,count;
trie tr[100010];
new_node()
{
        int i;
    count++;
    tr[count].flag = 0;
    for(i = 0;i < 10;i++)
        tr[count].next[i] = -1;
}
int insert(char *s,trie *root )
{
    int flag,shorts,ct,i;
    flag = 1;
    shorts = 1;
    ct = 0;
    for(i = 0;s[i] != '\0';i++)
    {
        if(root[ct].next[s[i]-'0'] == -1)
        {
            new_node();
            shorts = 0;
            root[ct].next[s[i]-'0'] = count;
        }
        ct = root[ct].next[s[i]-'0'];
        if(root[ct].flag)
            flag = 0;
    }
    root[ct].flag = 1;
    if(shorts)
        flag = 0;
    return flag;
}

int main()
{
    char s[100];
    int t,n,flag,i;
    scanf("%d",&t);
    while(t--)
    {
        count = -1;
        new_node();
        scanf("%d%*c",&n);
        leap = 1;
        while(n--)
        {
            gets(s);
            flag = 1;
            flag = insert(s,tr);
            if(!flag)
                leap = 0;
        }
        if(leap)
            puts("YES");
        else
            puts("NO");
    }

    return 0;
}

这个是动态的，结果跟某人说的一样，超时。毕竟动态的省空间，静态省时间嘛~

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<string.h>
using namespace std;
typedef struct node
{
    int flag;
    struct node *next[11];
    node()
    {
        flag = 0;
        memset(next,NULL,sizeof(next));
    }
}trie;
int leap;

int insert(char *s,trie *root )
{
    int i = 0;
    trie *p;
    int flag,shorts;
    flag = 1;
    shorts = 1;
    for(i = 0;s[i] != '\0';i++)
    {
        if(root->next[s[i]-'0'] == NULL)
        {
            root->next[s[i]-'0'] = new node;
            shorts = 0;
        }
        root = root->next[s[i]-'0'];
        if(root->flag)
            flag = 0;
    }
    root->flag = 1;
    if(shorts)
        flag = 0;
    return flag;
}

int main()
{
    char s[100];
    int t,n,flag,i;
    scanf("%d",&t);
    while(t--)
    {
        trie *root;
        root = new node;
        scanf("%d%*c",&n);
        leap = 1;
        while(n--)
        {
            gets(s);
            flag = 1;
            flag = insert(s,root);
            if(!flag)
                leap = 0;
        }
        if(leap)
            puts("YES");
        else
            puts("NO");
    }

    return 0;
}