Codeforces 558E A Simple Task

Discription

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.

Example

Input

10 5abacdabcda7 10 05 8 11 4 03 6 07 10 1

Output

cbcaaaabdd

Input

10 1agjucbvdfk1 10 1

Output

abcdfgjkuv

Note

First sample test explanation:

大概是第一次坐到沙发hhh

可以发现只有26个小写英文，那么就不难了，就是一个常数*26的带标记线段树。。。

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<algorithm>
#define ll long long
#define maxn 100005
using namespace std;
struct node{
	int sum[28];
	int tag;

	node operator +(const node& u)const{
		node r; r.tag=0;
		for(int i=1;i<27;i++) r.sum[i]=sum[i]+u.sum[i];
		return r;
	}
}a[maxn<<2|1];
int le,ri,opt,w;
int n,m,v,num[30];
char s[maxn];

void build(int o,int l,int r){
	if(l==r){
		a[o].sum[s[l]-‘a‘+1]++;
		return;
	}

	int mid=l+r>>1,lc=o<<1,rc=(o<<1)|1;
	build(lc,l,mid),build(rc,mid+1,r);
	a[o]=a[lc]+a[rc];
}

inline void tol(int o,int tmp,int len){
	memset(a[o].sum,0,sizeof(a[o].sum));
	a[o].sum[tmp]=len;
	a[o].tag=tmp;
}

inline void pushdown(int o,int l,int r){
	if(a[o].tag){
		int mid=l+r>>1,lc=o<<1,rc=(o<<1)|1;
		tol(lc,a[o].tag,mid-l+1);
		tol(rc,a[o].tag,r-mid);
		a[o].tag=0;
	}
}

void update(int o,int l,int r){
	if(l>=le&&r<=ri){
		tol(o,v,r-l+1);
		return;
	}
	pushdown(o,l,r);
	int mid=l+r>>1,lc=o<<1,rc=(o<<1)|1;
	if(le<=mid) update(lc,l,mid);
	if(ri>mid) update(rc,mid+1,r);
	a[o]=a[lc]+a[rc];
}

int query(int o,int l,int r){
	if(l>=le&&r<=ri) return a[o].sum[v];
	pushdown(o,l,r);
	int mid=l+r>>1,lc=o<<1,rc=(o<<1)|1,an=0;
	if(le<=mid) an+=query(lc,l,mid);
	if(ri>mid) an+=query(rc,mid+1,r);
	return an;
}

void print(int o,int l,int r){
	if(l==r){
		for(int i=1;i<27;i++) if(a[o].sum[i]){
			putchar(i-1+‘a‘);
			break;
		}
		return;
	}
	pushdown(o,l,r);
	int mid=l+r>>1,lc=o<<1,rc=(o<<1)|1;
	print(lc,l,mid);
	print(rc,mid+1,r);
}

int main(){
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	build(1,1,n);
	while(m--){
		scanf("%d%d%d",&le,&ri,&opt);
		memset(num,0,sizeof(num));
		for(v=1;v<=26;v++) num[v]=query(1,1,n);
		if(opt){
			int pre=le,nxt;
			for(int i=1;i<=26;i++) if(num[i]){
				nxt=pre+num[i]-1,v=i;
				le=pre,ri=nxt;
				update(1,1,n);
				pre=nxt+1;
			}
		}
		else{
			int pre=le,nxt;
			for(int i=26;i;i--) if(num[i]){
				nxt=pre+num[i]-1,v=i;
				le=pre,ri=nxt;
				update(1,1,n);
				pre=nxt+1;
			}
		}
//		print(1,1,n);
	}

	print(1,1,n);
	return 0;
}

原文地址：https://www.cnblogs.com/JYYHH/p/8407093.html