【codeforces 550A】Two Substrings

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given string s. Your task is to determine if the given string s contains two non-overlapping substrings “AB” and “BA” (the substrings can go in any order).

Input
The only line of input contains a string s of length between 1 and 105 consisting of uppercase Latin letters.

Output
Print “YES” (without the quotes), if string s contains two non-overlapping substrings “AB” and “BA”, and “NO” otherwise.

Examples
input
ABA
output
NO
input
BACFAB
output
YES
input
AXBYBXA
output
NO
Note
In the first sample test, despite the fact that there are substrings “AB” and “BA”, their occurrences overlap, so the answer is “NO”.

In the second sample test there are the following occurrences of the substrings: BACFAB.

In the third sample test there is no substring “AB” nor substring “BA”.

【题目链接】:http://codeforces.com/contest/550/problem/A

【题解】

把所有的ABABABA…以及BABA…这两种形式的子串“整个”提出来;
把它们的长度分别放在两个vector,ab以及ba中。
如果最后ab和ba两个vecto的size都大于0,则yes
如果ab或ba中最大的数字(也即ABAB..的长度或BABAB…的长度)大于等于5
则yes(因为ABABA中可以找到AB和BA,BABAB中也可以找到BAAB,且这是所需的最短的串);
又或者,ab或ba中有一个大小>=3,还有一个大小>=2,则也可行
比如ABTABA
ab最后有两个元素分别为2和3;
可以看到能够找到AB和BA.
BATBAB同理
也能找到BA和AB;
其他情况NO;

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

//const int MAXN = x;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

const int MAXN = 1e5+100;

string s;
vector <int> ab,ba;
int ab3,ba3,ab5,ba5;

int main()
{
    //freopen("F:\rush.txt","r",stdin);
    cin >> s;
    s = ' '+s;
    int len = s.size()-1;
    rep1(i,1,len)
        {
            if (s[i]=='A' && i+1<=len && s[i+1]=='B')
            {
                int r = i;
                while ((r+1<=len) && (s[r+1] == 'A' || s[r+1]=='B') && s[r+1]!=s[r])
                    r++;
                int le = r-i+1;
                if (le>=3)
                {
                    ab3++;
                    if (le>=5)
                        ab5++;
                }
                ab.pb(le);
                i = r;
            }
            else
                if (s[i]=='B' && i+1<=len && s[i+1]=='A')
                {
                    int r = i;
                    while ((r+1<=len) && (s[r+1] == 'A' || s[r+1]=='B') && s[r+1]!=s[r])
                        r++;
                    int le = r-i+1;
                    ba.pb(le);
                    if (le>=3)
                    {
                        ba3++;
                        if (le>=5)
                            ba5++;
                    }
                    i = r;
                }
        }
    int lenab = ab.size(),lenba = ba.size();
    if (lenab>0 && lenba >0)
    {
        puts("YES");
        return 0;
    }
    if (ab5||ba5)
    {
        puts("YES");
        return 0;
    }
    if (lenab >1 && ab3)
    {
        puts("YES");
        return 0;
    }
    if (lenba >1 && ba3)
    {
        puts("YES");
        return 0;
    }
    puts("NO");
    return 0;
}