Codeforces Round #312 (Div. 2) E. A Simple Task 线段树+计数排序

题目链接：

http://codeforces.com/problemset/problem/558/E

E. A Simple Task

time limit per test5 seconds
memory limit per test512 megabytes #### 问题描述 > 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.

输入

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, ).

输出

Output one line, the string S after applying the queries.

样例输入

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

样例输出

cbcaaaabdd

题意

给你一个长度为n的字符串（只有小写字母），m个更新操作（l,r,type）要么对区间里面的字符串升序排，要么降序排，问m个更新之后最后的字符串是什么

题解

首先，只有26个字母，可以考虑计数排序，其次，当然是要用线段树来维护了，可以开26颗线段树，支持区间更新，区间查询。
对于更新区间先查询一下各个字母多少个，然后再按排完序的结果，26个字母（对应26颗线段树）全部干一发区间更新。

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef __int64 LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1e5+10;

int sumv[26][maxn<<2],setv[26][maxn<<2];
int n,m;
char str[maxn];

void maintain(int id,int o) {
    sumv[id][o]=sumv[id][lson]+sumv[id][rson];
}

void pushdown(int id,int o,int l,int r) {
    if(setv[id][o]<0) return;
    sumv[id][lson]=(mid-l+1)*setv[id][o];
    sumv[id][rson]=(r-mid)*setv[id][o];
    setv[id][lson]=setv[id][rson]=setv[id][o];
    setv[id][o]=-1;
}

void build(int id,int o,int l,int r) {
    if(l==r) {
        int idx=str[l]-'a';
        if(id==idx) sumv[id][o]=1;
        else sumv[id][o]=0;
    } else {
        build(id,lson,l,mid);
        build(id,rson,mid+1,r);
        maintain(id,o);
    }
}

int ul,ur,uv;
void update(int id,int o,int l,int r) {
    if(ul<=l&&r<=ur) {
        sumv[id][o]=(r-l+1)*uv;
        setv[id][o]=uv;
    } else {
        pushdown(id,o,l,r);
        if(ul<=mid) update(id,lson,l,mid);
        if(ur>mid) update(id,rson,mid+1,r);
        maintain(id,o);
    }
}

int ql,qr,qv;
void query(int id,int o,int l,int r) {
    if(ql<=l&&r<=qr) {
        qv+=sumv[id][o];
    } else {
        pushdown(id,o,l,r);
        if(ql<=mid) query(id,lson,l,mid);
        if(qr>mid) query(id,rson,mid+1,r);
        maintain(id,o);
    }
}

void init() {
    clr(sumv,0);
    clr(setv,-1);
}

int main() {
    init();
    scf("%d%d",&n,&m);
    scf("%s",str+1);

    rep(i,0,26) build(i,1,1,n);

    rep(i,0,m) {
        int l,r,type;
        scf("%d%d%d",&l,&r,&type);
        int lef=l;
        if(type==1) {
            rep(i,0,26) {
                ql=l,qr=r,qv=0;
                query(i,1,1,n);

                ul=l,ur=r,uv=0;
                update(i,1,1,n);

                ul=lef,ur=lef+qv-1,uv=1;
                if(ul<=ur) update(i,1,1,n);
                lef+=qv;
            }
        } else {
            for(int i=25; i>=0; i--) {
                ql=l,qr=r,qv=0;
                query(i,1,1,n);

                ul=l,ur=r,uv=0;
                update(i,1,1,n);

                ul=lef,ur=lef+qv-1,uv=1;
                if(ul<=ur){
//                    prf("(%d,%d),id:%d
",ul,ur,i);
                    update(i,1,1,n);
                }
                lef+=qv;
            }
        }
    }

    for(int i=1;i<=n;i++){
        rep(j,0,26){
            ql=i,qr=i,qv=0;
            query(j,1,1,n);
            if(qv>0){
                prf("%c",'a'+j);
                break;
            }
        }
    }
    puts("");

    return 0;
}

//end-----------------------------------------------------------------------

/*
10 1
abacdabcda
7 10 0
*/

Notes

以后需要建多颗线段树的时候最好把它们所有的操作都独立开，不要放在一起！因为放在一起反而有可能变麻烦。