CF 558E (A simple task-counting sort + line tree)

E. A simple Task time limit per test 5 seconds memory limit per test megabytes input standard input Output standard OU Tput

This task was very simple. Given A string S of length n and q queries each query are on the format I J K which means sort the substring consisting of The characters from I to J in non-decreasing order if k = 1 or in non-increasing order if k = 0.

Output the final string after applying the queries. Input

The first line would contain, integers n, q (1≤n≤105, 0≤q≤50), the length of the string and the number of Qu Eries respectively.

Next line contains a string S itself. It contains only lowercase 中文版 letters.

Next q lines would contain three integers each I, j, K (1≤i≤j≤n,). Output

Output one line, the string S after applying the queries. Sample Test (s) input

5
abacdabcda
7 0
5 8 1
1 4 0
3 6 0
7 10 1

Output

Cbcaaaabdd

Input

1
AGJUCBVDFK
1 10 1

Output

Abcdfgjkuv

Note

First sample test explanation:

Counting sequencing, the key is how to maintain

Calculation found 26 tree segment, the complexity of O (M*26LOGN)

#include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #include < functional> #include <iostream> #include <cmath> #include <cctype> #include <ctime> using
namespace Std; #define for (I,n) for (int. i=1;i<=n;i++) #define FORK (I,k,n) for (int. i=k;i<=n;i++) #define REP (i,n) for (int i=0;i< n;i++) #define ForD (I,n) for (int. i=n;i;i--) #define REPD (I,n) for (int. i=n;i>=0;i--) #define FORP (x) for (int p=pre[x];p; P=NEXT[P]) #define FORPITER (x) for (int &p=iter[x];p; p=next[p]) #define LSON (o<<1) #define Rson ((o<<1) +
1) #define MEM (a) memset (A,0,sizeof (a));
#define MEMI (a) memset (A,127,sizeof (a));
#define MEMI (a) memset (A,128,sizeof (a));
#define INF (2139062143) #define F (100000007) #define MAXN (400000+10) #define MAXQ (50000+10) typedef long Long LL; LL Mul (ll A,ll b) {return (a*b)%F;} ll Add (ll A,ll b) {return (a+b)%F;} ll Sub (ll A,ll b) {return (a-b+ (-a)/f*f+f)%F; void UPD (ll &a,ll b) {a= (a%f+b%f)%F;}
int n,q;
int L,r,v,_ans;
	Class stree{Public:int MARK[MAXN],TREE[MAXN]; Stree () {mem (Mark) mem (tree)} void pushdown (int o,int l,int R) {if (l<r) if (Mark[o]) mark[lson]=mark[rson]=m
	ark[o],mark[o]=0;
		} void maintain (int o,int l,int R) {if (Mark[o]) tree[o]= (r-l+1) * (mark[o]-1);
	else if (l<r) Tree[o]=tree[lson]+tree[rson];
			} void Set (int o,int l,int R) {if (l<=l&&r<=r) {mark[o]=v;
			Maintain (O,L,R);
		Return
		} pushdown (O,l,r);
		int m= (L+R) >>1;
		if (l<=m) set (LSON,L,M); else maintain (lson,l,m);
		if (m<r) set (RSON,M+1,R); else maintain (RSON,M+1,R);
	Maintain (O,L,R);
			} void get_sum (int o,int l,int R) {if (l<=l&&r<=r) {_ans+=tree[o]; 
		Return
			} if (Mark[o]) {_ans+= (min (r,r)-max (l,l) +1) * (mark[o]-1);
		Return
		} int m= (L+R) >>1;
		if (l<=m) get_sum (lson,l,m);
	if (m<r) get_sum (rson,m+1,r);
}}s[26];

Char S[MAXN];

int cnt[26]={0}; int main () {//Freopen ("E.in "," R ", stdin);
	
	Freopen (". Out", "w", stdout);
	scanf ("%d%d%s", &n,&q,s);
		Rep (i,n) {l=r=i+1,v=2;
	s[s[i]-' A '].set (1,1,n);
		} for (qcase,q) {int b;
		
		scanf ("%d%d%d", &l,&r,&b);
		V=1;
			Rep (i,26) {_ans=0;
			S[i].get_sum (1,1,n);
			Cnt[i]=_ans;
		S[i].set (1,1,n);
		} v=2;
				if (b)//increasing {REP (i,26) {if (!cnt[i]) continue; 
				R=l+cnt[i]-1;
				S[i].set (1,1,n);
			l=r+1;
				}} else {Rep (i,26) {if (!cnt[i]) continue; 
				l=r-cnt[i]+1;
				S[i].set (1,1,n);
			R=L-1;
			}}} for (I,n) {Rep (j,26) {l=r=i;_ans=0;
			S[j].get_sum (1,1,n);
				if (_ans) {s[i-1]= ' a ' +j;
			Break
	
	}}} printf ("%s\n", s);
return 0;
 }