E. String Reversal(逆序对)

Posted 2022-02-09 thusloop

String Reversal
题意：给出字符串S 求 交换相邻字符后变成 S的逆序 的最小操作。
思路：给S的逆序每个字符从小到大赋值 再 对应到S中 即求S的逆序对个数
比如 S:aaaza 的逆序 T:azaaa t[]=1,2,3,3,3. s[]=1,3,3,2,3

//#pragma GCC optimize(2)
//#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
#define int long long
#define fi first
#define se second
#define pb push_back
#define pii pair<int,int>
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
const int inf=8e18;
const int maxn=2e5+100;
char a[maxn],b[maxn];
int d[maxn],e[maxn],cnt[30];
int c[maxn];
int lowbit(int x)

	return x&(-x);

void update(int x)

	while(x<maxn)
	
		c[x]++;
		x+=lowbit(x);
	

int ask(int x)

	int ans=0;
	while(x)
	
		ans+=c[x];
		x-=lowbit(x);
	
	return ans;

signed main()

	vector<int>g[30];
	int n;
	cin>>n;
	cin>>(a+1);
	for(int i=1;i<=n;i++)
	
		b[i]=a[i];
		g[a[i]-'a'].pb(i);
	
	reverse(b+1,b+n+1);
	int mx=0;
	for(int i=1;i<=n;i++)
	
		if(b[i]!=b[i-1]) e[i]=++mx;
		else e[i]=mx;
	
	for(int i=1;i<=n;i++)
	
		d[g[b[i]-'a'][cnt[b[i]-'a']]]=e[i];
		cnt[b[i]-'a']++;
	
	int ans=0;
	for(int i=1;i<=n;i++)
	
		ans+=i-ask(d[i])-1;
		update(d[i]);
	
	cout<<ans<<"\\n";