Sicily 1426. Phone List

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判断一系列字符串中，是否存在一个字符串是另一个字符串的前缀。可以按字典序排一个序，然后前后两个对比一下就行；虽然此法比较耗时，但胜在简单易懂，空间消耗小，代码短。

Run Time: 0.27sec

Run Memory: 708KB

Code Length: 601Bytes

Submit Time: 2011-06-23 20:31:06

// Problem#: 1426// Submission#: 823325// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <iostream>#include <string>#include <algorithm>using namespace std;int main(){    string s[ 10000 ];    int t, n;    int i;    bool find;    cin >> t;    while ( t-- ) {        cin >> n;        for ( i = 0; i < n; i++ )            cin >> s[ i ];        sort( s, s + n );        find = false;        for ( i = 1; i < n; i++ ) {            if ( s[ i ].find( s[ i - 1 ] ) != string::npos ) {                find = true;                break;            }        }        cout << ( find ? "NO": "YES" ) << endl;    }    return 0;}