Codeforces Round #723 (Div. 2)D. Kill Anton 贪心，逆序对个数

Posted 2021-06-21 小哈里

problem

D. Kill Anton
time limit per test2 seconds
memory limit per test512 megabytes
inputstandard input
outputstandard output
After rejecting 10100 data structure problems, Errorgorn is very angry at Anton and decided to kill him.

Anton’s DNA can be represented as a string a which only contains the characters “ANTON” (there are only 4 distinct characters).

Errorgorn can change Anton’s DNA into string b which must be a permutation of a. However, Anton’s body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton’s body is smart and will use the minimum number of moves.

To maximize the chance of Anton dying, Errorgorn wants to change Anton’s DNA the string that maximizes the time for Anton’s body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?

Input
The first line of input contains a single integer t (1≤t≤100000) — the number of testcases.

The first and only line of each testcase contains 1 string a (1≤|a|≤100000). a consists of only the characters “A”, “N”, “O” and “T”.

It is guaranteed that the sum of |a| over all testcases does not exceed 100000.

Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.

Example
inputCopy
4
ANTON
NAAN
AAAAAA
OAANTTON
outputCopy
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton’s body to transform NNOTA to ANTON:

NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON.

Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.

For the second testcase, it takes 2 seconds for Anton’s body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.

solution

//相同字母一定是连着的
//顺序的话，4!枚举，操作次数是编号后逆序对进行比较
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
int z[50];
vector<int>id[50], vc;
LL nxd = 0;
void MergeSort(int l, int r){
    if(l >= r)return ;
    int m = l+r>>1;
    MergeSort(l,m);
    MergeSort(m+1,r);
    int i = l, j = m+1;
    int t[r-l+1], k = 0;
    while(i<=m && j<=r){
        if(vc[i]<=vc[j])t[k++]=vc[i++];
        else{
            t[k++] = vc[j++];
            nxd += m-i+1;//加上剩余元素个数
        }
    }
    while(i<=m)t[k++]=vc[i++];
    while(j<=r)t[k++]=vc[j++];
    for(i=l, k=0; i <= r; i++,k++)vc[i]=t[k];
}
int main(){
	ios::sync_with_stdio(false);
	int T;  cin>>T;
	while(T--){
		memset(z,0,sizeof(z));
		id['A'-'A'].clear();
		id['N'-'A'].clear();
		id['O'-'A'].clear();
		id['T'-'A'].clear();
		string s;  cin>>s;
		for(int i = 0; i < s.size(); i++){
			z[s[i]-'A']++;
			id[s[i]-'A'].push_back(i+1);
		}
		int r[5] = {0,'A','N','O','T'};
		LL ans = 0; string as;
		do{
			//vector<int>vc;
			vc.clear();
			nxd = 0;
			for(int i = 1; i <= 4; i++){
				vc.insert(vc.end(),id[r[i]-'A'].begin(),id[r[i]-'A'].end());
			}
			MergeSort(0,vc.size()-1);
			if(nxd>ans){
				ans = nxd;
				as = "";
				for(int i = 1; i <= 4; i++)
					as += string(z[r[i]-'A'],(char)r[i]);
			}
		}while(next_permutation(r+1,r+5));
		//cout<<ans<<"\\n";
		if(ans!=0)cout<<as<<"\\n";
		else cout<<s<<"\\n";
	}
	return 0;
}