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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.
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 one line, the string S after applying the queries.
10 5 abacdabcda 7 10 0 5 8 1 1 4 0 3 6 0 7 10 1
cbcaaaabdd
10 1 agjucbvdfk 1 10 1
abcdfgjkuv
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;
}
版权声明：本文为博主原创文章，未经博主允许不得转载。
CF 558E(A Simple Task-计数排序+线段树)
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原文地址：http://blog.csdn.net/nike0good/article/details/47067915
