POJ 3630 Phone List(字串字首重複)題解
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 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
Sample Output
NO YES
思路:
樹狀陣列超時。這裡給每一串先排序,然後比較相鄰的兩個有沒有字首重複就可以了,有點厲害(所以為什麼他會出現在樹狀陣列專題呢)。
程式碼:
#include<cstdio> #include<cstring> #include<cstdlib> #include<queue> #include<cmath> #include<string> #include<stack> #include<set> #include<map> #include<vector> #include<iostream> #include<algorithm> #include<sstream> #define ll long long const int N=10005; const int INF=1e9; using namespace std; char s[N][15]; int cmp(const void *a,const void *b){ return strcmp((char *)a,(char *)b); } int main(){ int flag,t,n; scanf("%d",&t); while(t--){ scanf("%d",&n); for(int i=0;i<n;i++){ scanf("%s",s[i]); } qsort(s,n,sizeof(s[0]),cmp); flag=0; for(int i=0;i<n-1;i++){ if(strncmp(s[i],s[i+1],strlen(s[i]))==0){ flag=1; break; } } if(flag) printf("NO\n"); else printf("YES\n"); } return 0; }