C - AB Substrings

最新推荐文章于 2020-04-26 22:17:20 发布

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本文探讨了一种算法，旨在通过重新排列并连接一组字符串，以最大化特定子串“AB”的出现次数。该方法首先分析每个字符串的首尾字符，然后根据特定规则进行连接，以确保目标子串的最大化出现。

Problem Statement
Snuke has N strings. The i-th string is si.

Let us concatenate these strings into one string after arranging them in some order. Find the maximum possible number of occurrences of AB in the resulting string.

Constraints1≤N≤1042≤|si|≤10si consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:

N s1。。。sN

Output
Print the answer.

Sample Input 1
Copy
3
ABCA
XBAZ
BAD
Sample Output 1
Copy
2
For example, if we concatenate ABCA, BAD and XBAZ in this order, the resulting string ABCABADXBAZ has two occurrences of AB.

Sample Input 2
Copy
9
BEWPVCRWH
ZZNQYIJX
BAVREA
PA
HJMYITEOX
BCJHMRMNK
BP
QVFABZ
PRGKSPUNA
Sample Output 2
Copy
4
Sample Input 3
Copy
7
RABYBBE
JOZ
BMHQUVA
BPA
ISU
MCMABAOBHZ
SZMEHMA
Sample Output 3
Copy
4

这个思路叭，可以说是最蠢的了，就是把字符串按规则连起来，暴力
容易超时

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<ctime>
#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<string>
#define ll long long
#define dd double
using namespace std;
 
int main(){
	string s[10005];
	string st="";
	ll a[10005][3]={0};
	memset(a,0,sizeof(a));
	ll n;cin>>n;
	for(ll i=0;i<n;i++){
		cin>>s[i];
		if(s[i][0]=='B'){
			a[i][0]++;
		}
		if(s[i][s[i].size()-1]=='A'){
			a[i][1]++;
		}
	}
	ll flag=0;
	for(ll i=0;i<n;i++){
		if(a[i][0]==0&&a[i][1]!=0){
			flag=1;
			st+=s[i];
			a[i][2]=-1;
			break;
		}
	}
	ll q=0;
	if(flag==1){
		for(ll i=0;i<n;i++){
			if(a[i][2]==-1){
				continue;
			}
			else{
				if(a[i][0]!=0&&a[i][1]!=0){
					st+=s[i];
					a[i][2]=-1;
				}
			}
		}
		for(ll i=0;i<n;i++){
			if(a[i][2]==-1){
				continue;
			}
			else{
				if(a[i][0]!=0){
					st+=s[i];
					a[i][2]=-1;
					break;
				}
			}
		}
		for(ll i=0;i<n;i++){
			if(a[i][2]==-1){
				continue;
			}
			else{
				if(q%2==0){
					if(a[i][1]!=0){
						st+=s[i];
						a[i][2]=-1;
						i=-1;
						q++;
					}
				}
				else{
					if(a[i][0]!=0){
						st+=s[i];
						a[i][2]=-1;
						i=-1;
						q++;
					}
				}
			}
		}
		for(ll i=0;i<n;i++){
			if(a[i][2]!=-1){
				st+=s[i];
				a[i][2]=-1;
			}
		}
	}
	if(flag==0){
		for(ll i=0;i<n;i++){
			if(a[i][1]!=0&&a[i][2]!=-1){
				st+=s[i];
				a[i][2]=-1;
				break;
			}
		}
		for(ll i=0;i<n;i++){
			if(a[i][2]==-1){
				continue;
			}
			else{
				if(a[i][0]!=0&&a[i][1]!=0){
					st+=s[i];
					a[i][2]=-1;
				}
			}
		}
		for(ll i=0;i<n;i++){
			if(a[i][2]==-1){
				continue;
			}
			else{
				if(a[i][0]!=0){
					st+=s[i];
					a[i][2]=-1;
				}
			}
		}
		for(ll i=0;i<n;i++){
			if(a[i][2]!=-1){
				st+=s[i];
				a[i][2]=-1;
			}
		}
	}
	ll count = 0;
	for(ll i=0;i<st.size()-1;i++){
		if(st[i]=='A'&&st[i+1]=='B'){
			count++;
		}
	}
	cout<<count<<endl;
}