ACM-ICPC国际大学生程序设计竞赛北京赛区(2015)网络练习赛 题目4 : Beautiful String

We say a string is beautiful if it has the equal amount of 3 or more continuous letters (in increasing order.)

Here are some example of valid beautiful strings: "abc", "cde", "aabbcc", "aaabbbccc".

Here are some example of invalid beautiful strings: "abd", "cba", "aabbc", "zab".

Given a string of alphabets containing only lowercase alphabets (a-z), output "YES" if the string contains a beautiful sub-string, otherwise output "NO".

输入

The first line contains an integer number between 1 and 10, indicating how many test cases are followed.

For each test case: First line is the number of letters in the string; Second line is the string. String length is less than 10MB.

输出

For each test case, output a single line "YES"/"NO" to tell if the string contains a beautiful sub-string.

提示

Huge input. Slow IO method such as Scanner in Java may get TLE.

样例输入

4
3
abc
4
aaab
6
abccde
3
abb

样例输出

YES
NO
YES
NO

#include<iostream>
#include<vector>
#include<string>
using  namespace std;

int main() {
    int n;
    cin>>n;
    while( n-- ) {
        vector<char> v;
        int num;
        cin>>num;
        string str;
        cin>>str;
        vector<char> v1;
        vector<int> v2;
        v1.push_back(str[0]);
        v2.push_back(1);
        int flag = 1;
    
        for( int i = 1; i < str.size(); i++ ) {
            if( str[i] == str[i-1] ) {
                int temp = *(v2.end()-1);
                temp++;
                v2.pop_back();
                v2.push_back(temp);
                if( v2[1] == v2[2] && v2[0] >= v2[1] && v1[2] - v1[1] == 1 && v1[1] - v1[0] == 1) {
                        cout<<"YES"<<endl;
                        flag = 0;
                        break;
                    }
            }
            else {
                if( v1.size() < 3 ) {
                    v1.push_back(str[i]);
                    v2.push_back(1);
                    if( v2[1] == v2[2] && v2[0] >= v2[1] && v1[2] - v1[1] == 1 && v1[1] - v1[0] == 1) {
                        cout<<"YES"<<endl;
                        flag = 0;
                        break;
                    }
                }
                else {                    
                        v1.erase(v1.begin());
                        v2.erase(v2.begin());
                        v1.push_back(str[i]);
                        v2.push_back(1);
                        if( v2[1] == v2[2] && v2[0] >= v2[1] && v1[2] - v1[1] == 1 && v1[1] - v1[0] == 1) {
                        cout<<"YES"<<endl;
                        flag = 0;
                        break;
                    }
                    
                }
            }
        }    
        if( flag ) 
            cout<<"NO"<<endl;    
    }
    return 0;
}