Description
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 911Alice 97 625 999Bob 91 12 54 26In 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 98346Sample Output
NO YES这道题是一道tried树的模板题，下面是程序：
#include<stdio.h> #include<algorithm> #include<string.h> #include<iostream> using namespace std; struct node{ int x,next[10]; }m[100005]; struct num{ int l; char a[15]; bool operator <(num p)const { return l>p.l; } }x[10005]; int k; void add(int u,int w){ m[u].next[w]=++k; } int dfs(int root,int h,num &p){ int &q=m[root].next[p.a[h]-'0']; if(h==p.l-1){ if(q==-1){ add(root,p.a[h]-'0'); return -1; } else{ return q; } } if(q!=-1){ return dfs(q,h+1,p); } add(root,p.a[h]-'0'); dfs(q,h+1,p); return -1; } bool work(){ k=0; memset(m,-1,sizeof(m)); int n,i; scanf("%d",&n); for(i=1;i<=n;i++){ scanf("%s",x[i].a); x[i].l=strlen(x[i].a); } sort(x+1,x+n+1); for(i=1;i<=n;i++){ if(dfs(0,0,x[i])!=-1){ return 0; } } return 1; } int main(){ int t; scanf("%d",&t); while(t--){ printf("%s\n",work()?"YES":"NO"); } return 0; } 注：要注意常数。
