CF 558E(A Simple Task-计数排序+线段树)

E. A Simple Task

time limit per test

5 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

This task is very simple. Given a string S of length n and q queries
each query is 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 will contain two integers n,?q (1?≤?n?≤?105, 0?≤?q?≤?50?000),
the length of the string and the number of queries respectively.

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

Next q lines will 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

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

output

cbcaaaabdd

input

10 1
agjucbvdfk
1 10 1

output

abcdfgjkuv

Note

First sample test explanation:

计数排序，关键是如何维护

计算发现建26棵线段树，复杂度为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-b)/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]=mark[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;
}

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