线段树统计和维护某一区间内的字母个数。。
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A boy named Leo doesn't miss a single TorCoder contest round. On the last TorCoder round number 100666 Leo stumbled over the following problem. He was given a string s, consisting of n lowercase English letters, and m queries. Each query is characterised by a pair of integers li, ri (1 ≤ li ≤ ri ≤ n).
We'll consider the letters in the string numbered from 1 to n from left to right, that is, s = s1s2... sn.
After each query he must swap letters with indexes from li to ri inclusive in string s so as to make substring (li, ri) a palindrome. If there are multiple such letter permutations, you should choose the one where string (li, ri) will be lexicographically minimum. If no such permutation exists, you should ignore the query (that is, not change string s).
Everybody knows that on TorCoder rounds input line and array size limits never exceed 60, so Leo solved this problem easily. Your task is to solve the problem on a little bit larger limits. Given string s and m queries, print the string that results after applying all m queries to strings.
The first input line contains two integers n and m (1 ≤ n, m ≤ 105) — the string length and the number of the queries.
The second line contains string s, consisting of n lowercase Latin letters.
Each of the next m lines contains a pair of integers li, ri (1 ≤ li ≤ ri ≤ n) — a query to apply to the string.
In a single line print the result of applying m queries to string s. Print the queries in the order in which they are given in the input.
7 2 aabcbaa 1 3 5 7
abacaba
3 2 abc 1 2 2 3
abc
A substring (li, ri) 1 ≤ li ≤ ri ≤ n) of string s = s1s2... sn of length n is a sequence of characters slisli + 1...sri.
A string is a palindrome, if it reads the same from left to right and from right to left.
String x1x2... xp is lexicographically smaller than string y1y2... yq, if either p < q and x1 = y1, x2 = y2, ... , xp = yp, or exists such number r(r < p, r < q), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 < yr + 1.
/* ***********************************************
Author :CKboss
Created Time :2015年07月20日 星期一 07时29分48秒
File Name :CF240F_2.cpp
************************************************ */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
using namespace std;
const int maxn=100100;
int n,m;
int sum[28][maxn<<2];
int add[maxn<<2]; /// lazy
char str[maxn];
int cs[28];
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
void cls(int rt)
{
add[rt]=-1;
for(int i=0;i<26;i++) sum[i][rt]=0;
}
void push_up(int rt)
{
for(int i=0;i<26;i++)
sum[i][rt]=sum[i][rt<<1]+sum[i][rt<<1|1];
}
void push_down(int l,int r,int rt)
{
if(add[rt]!=-1)
{
int m=(l+r)/2;
int id = add[rt];
cls(rt<<1); cls(rt<<1|1);
add[rt<<1]=add[rt<<1|1]=add[rt];
sum[id][rt<<1]=m-l+1; sum[id][rt<<1|1]=r-m;
add[rt]=-1;
}
}
void build(int l,int r,int rt)
{
add[rt]=-1;
if(l==r)
{
int id=str[l]-'a';
sum[id][rt]++; add[rt]=id;
return ;
}
int m=(l+r)/2;
build(lson); build(rson);
push_up(rt);
}
void query(int L,int R,int l,int r,int rt)
{
if(L<=l&&r<=R)
{
for(int i=0;i<26;i++) cs[i]+=sum[i][rt];
return ;
}
push_down(l,r,rt);
int m=(l+r)/2;
if(L<=m) query(L,R,lson);
if(R>m) query(L,R,rson);
push_up(rt);
}
void update(int id,int L,int R,int l,int r,int rt)
{
if(L<=l&&r<=R)
{
cls(rt);
add[rt]=id;
sum[id][rt]=r-l+1;
return ;
}
push_down(l,r,rt);
int m=(l+r)/2;
if(L<=m) update(id,L,R,lson);
if(R>m) update(id,L,R,rson);
push_up(rt);
}
char res[maxn];
void solve(int L,int R)
{
int op=-1,odd=0;
memset(cs,0,sizeof(cs));
query(L,R,1,n,1);
for(int i=0;i<26;i++)
{
if(cs[i]%2) { odd++; op=i; }
if(odd>1) return ;
}
int st=L,en=R;
int MID=L+(R-L+1)/2;
for(int i=0;i<26;i++)
{
if(cs[i])
{
if(cs[i]/2)
{
update(i,st,st+cs[i]/2-1,1,n,1);
update(i,en-cs[i]/2+1,en,1,n,1);
st=st+cs[i]/2;
en=en-cs[i]/2;
}
if(cs[i]%2==1)
{
update(i,MID,MID,1,n,1);
}
}
}
}
int main()
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
scanf("%d%d",&n,&m);
scanf("%s",str+1);
/// build
build(1,n,1);
int L,R;
while(m--)
{
scanf("%d%d",&L,&R); solve(L,R);
}
for(int i=1;i<=n;i++)
{
memset(cs,0,sizeof(cs));
query(i,i,1,n,1);
for(int j=0;j<26;j++)
if(cs[j])
{
putchar('a'+j); break;
}
}
putchar(10);
return 0;
}
