POJ 3630 Phone List 解题报告 字典树

POJ 3630 Phone List 解题报告 字典树

解题思路：建一个字典树再进行判断就行了。详情见代码，据说用malloc分配空间会超时？这里用的静态数组。调bug调了我半天，千万注意指针的初始化。

#include<iostream>
#include<math.h>
#include<iomanip>
#include<algorithm>
#include<iostream>
#include<math.h>
#include<iomanip>
#include<algorithm>
#include<queue>
#include<cstring>
#include<string>
#include<map>
#include<stack>
#include<stdio.h>
#include<cstdio>
#include<stdlib.h>
#include<fstream>
#include<iomanip>
#pragma warning(disable:4996)
#define INF 0x3f3f3f3f
#define ll long long
#define PI acos(-1.0)
const int N = 1000010;
const int maxn = 1e9;
using namespace std;
int cnt, n;
int judge;
struct node {
	int flag;
	node* next[10];
}a[100000];
void build(node* t, char str[])
{
	int len = strlen(str);
	for (int i = 0; i < len; i++)
	{
		if (t->next[str[i] - '0'] == NULL)
		{
			cnt++;
			node* tem = &a[cnt];
			for (int i = 0; i < 9; i++)
			{
				tem->next[i] = NULL;
			}
			if (i == len - 1)
				tem->flag = 2;//标记结尾
			else
				tem->flag = 1;//标记已到达过该节点
			t->next[str[i] - '0'] = tem;
			if (i == len - 1)
				return;//如果结尾处数字是新建的，直接返回
		}
		t = t->next[str[i] - '0'];
		if (t->flag == 2)//如果是之前某串数字的结尾，说明前缀重复了
			judge = 1;
		if (i == len - 1)//如果结尾处数字不是新建的，说明前缀重复了
			judge = 1;
	}
}

int main()
{
	int t;
	scanf("%d", &t);
	char ch[15] = "";
	while (t--)
	{
		judge = 0;
		memset(a, 0, sizeof(a));
		cnt = 0;
		scanf("%d", &n);
		node* temp = &a[0];//初始化头指针
		for (int i = 0; i < 9; i++)
		{
			temp->next[i] = NULL;
		}
		temp->flag = 0;
		for (int i = 0; i < n; i++)
		{
			scanf("%s", &ch);
			if (judge)
				continue;
			build(temp, ch);
			memset(ch, 0, sizeof(ch));
		}
		if (judge)
		{
			printf("NO\n");
		}
		else
		{
			printf("YES\n");
		}
	}
}