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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:
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
Sample Output
NO YES
Source
#include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> using namespace std; #define ll long long #define mod 1000000007 #define pi (4*atan(1.0)) const int N=1e5+10,M=5e6+10,inf=1e9+10; int a[N][27],sum[M],len; void init() { memset(a,0,sizeof(a)); memset(sum,0,sizeof(sum)); len=1; } int getnum(char a) { return a-‘0‘; } void insertt(char *aa) { int u=0,n=strlen(aa); for(int i=0; i<n; i++) { int num=getnum(aa[i]); if(!a[u][num]) { a[u][num]=len++; } u=a[u][num]; sum[u]++; } } int getans(char *aa) { int u=0,x=strlen(aa); for(int i=0; i<x; i++) { int num=getnum(aa[i]); if(!a[u][num]) return 0; u=a[u][num]; } return sum[u]; } char ch[N][15]; int main() { int x,y,z,i,t; int T; scanf("%d",&T); while(T--) { scanf("%d",&x); init(); for(i=0; i<x; i++) { scanf("%s",ch[i]); insertt(ch[i]); } int flag=1; for(i=0;i<x;i++) { if(getans(ch[i])!=1) { flag=0; break; } } if(flag) printf("YES\n"); else printf("NO\n"); } return 0; }
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原文地址：http://www.cnblogs.com/jhz033/p/5581096.html
