ZJU 2876 解题报告

Phone List

Time Limit: 5 Seconds Memory Limit: 32768 KB

Given a list of phone numbers, determine if it is consistent in the sense that no number is the prefix of another. Let's say the phone catalogue listed these numbers:

- Emergency 911

- Alice 97 625 999

- Bob 91 12 54 26

In this case, it's not possible to call Bob, because the central would direct your call to the emergency line as soon as you had dialled the first three digits of Bob's phone number. So this list would not be consistent.

Input

The first line of input gives a single integer, 1 <= t <= 40, the number of test cases. Each test case starts with n, the number of phone numbers, on a separate line, 1 <= n <= 10000.Then follows n lines with one unique phone number on each line. A phone number is a sequence of at most ten digits.

Output

For each test case, output "YES" if the list is consistent, or "NO" otherwise.

Sample Input

2
3
911
97625999
91125426
5
113
12340
123440
12345
98346

Output for Sample Input

NO
YES

Source: The 2007 Nordic Collegiate Programming Contest

从小到大排序之后,如果存在某一个电话号码是另一个的前缀,那么它一定会作为后一个电话号码的前缀.开始试图用trie做,可惜没成功,暂时还没找出原因.另一种比较取巧的办法是,把电话号码做为64位整数读入,这样花在排序上的时间会少一些,不过还是没成功,电话号码可能是由0打头的.

CODE

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
#include<vector>
using namespace std;
 
int main()
{
  int i, j, k, n;
  vector <string> v;
  char s[15];
  scanf("%d", &n);
  while (n--)
    {
      scanf("%d", &k);
      v.clear();
      j = k;
      while (j--)
        {
          scanf("%s", s);
          v.push_back(s);
        }
      sort(v.begin(), v.end());
      --k;
      for(i = 0; i < k; ++i)
        {
          if(v[i+1].find(v[i]) == 0)
            {
              puts("NO");
              goto out;
            }
        }
      puts("YES");
out:;
    }
  return 0;
}

以下两段代码分别使用trie和long long做的,不过都没成功.

#include<stdio.h>
#include<string.h>
 
struct trie
{
  trie *node[10];
  bool end;
  trie()
    {
      end = false;
      for(int i = 0; i < 10; ++i)
        node[i] = NULL;
    }
} *root;
 
bool insert(char *s)
{
  bool T = true;
  trie *p = root;
  int i, c;
  for(i = 0; s[i]; ++i)
    {
      c = s[i] - '0';
      if(p -> node[c] == NULL) p -> node[c] = new trie();
      p = p -> node[c];
      if(p -> end)
        {
          T = false;
          break;
        }
    }
  p -> end = true;
  return T;
}
 
void release(trie *p)
{
  for(int i = 0; i < 10; ++i)
    if(p -> node[i]) release(p -> node[i]);
  delete(p);
}
 
int main()
{
  int i, j, k, t;
  char s[15];
  scanf("%d", &t);
  while (t--)
    {
      root = new trie();
      scanf("%d", &k);
      bool T = false;
      while(k--)
        {
          scanf("%s", s);
          if(T == false && insert(s) == false)
            {
              printf("NO\n");
              T = true;
            }
        }
      if(T == false) printf("YES\n");
      release(root);
    }
  return 0;
}
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
 
int main()
{
  long long D[10005], tmp;
  int i, j, k, n;
  scanf("%d", &n);
  while(n--)
    {
      scanf("%d", &k);
      for(i = 0; i < k; ++i)
        {
          scanf("%lld", &tmp);
          D[i] = tmp;
        }
      sort(D,D+k);
      --k;
      for(i = 0; i < k; ++i)
        {
          tmp = D[i+1];
          while(tmp > D[i]) tmp /= 10;
          if(tmp == D[i])
            {
              puts("NO");
              goto out;
            }
        }
      puts("YES");
out:;
    }
  return 0;
}

trie写的那个代码如果有人能帮我找出原因,真是感激不尽!

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