111qqz的小窝

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16/10/5 20:05

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codeforces

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codeforces 240 F. TorCoder (线段树)

题目链接

题意：给一个仅由小写字母组成的字符串，然后m个操作，每个操作一个区间，要求把区间中排列成字典序最小的回文串，如果不能形成回文串，就忽略该操作。

思路：和上一道线段树优化计数排序的题目很像，几乎是一样的。

同样的，26棵线段树，每种字母对应一棵。

每次统计询问区间中每种字母的个数。

然后先判断是否能形成回文（奇数长度只有一个个数为奇数的，偶数长度不能出现个数为奇数的）

能的话重置区间，然后前后分别覆盖。

注意如果是奇数长度的话，记得先覆盖中间点。

需要注意这道题的输入输出方式不是标准的。。。而是要加文件。。。不然会MLE 1

然而Tle 19….Tle了好久。。。

2016-10-05-17-43-28-%e7%9a%84%e5%b1%8f%e5%b9%95%e6%88%aa%e5%9b%be

最后换了种线段树的写法就过了。。。

然而后面这种写法就一定好么。。。。好像也不是。。。

总之是个挺玄学的东西。。。。不管了。。。

C++
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/* ***********************************************
Author :111qqz
Created Time :2016年10月05日 星期三 04时00分51秒
File Name :code/cf/problem/240F.cpp
************************************************ */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <deque>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <bitset>
#include <bits/stdc++.h>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair
using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=1E5+1;
int tree[26][N*4];
int lazy[26][N*4];
int n,m;
char st[N];
int cnt[26];
int L,R;
inline bool scan_d(int &num)  
{
        char in;bool IsN=false;
        in=getchar();
        if(in==EOF) return false;
        while(in!='-'&&(in<'0'||in>'9')) in=getchar();
        if(in=='-'){ IsN=true;num=0;}
        else num=in-'0';
        while(in=getchar(),in>='0'&&in<='9'){
                num*=10,num+=in-'0';
        }
        if(IsN) num=-num;
        return true;
}
void PushUp(int rt,int id)
{
    tree[id][rt] = tree[id][rt<<1] + tree[id][rt<<1|1];
}
void PushDown(int l,int r,int rt,int id)
{
    if (lazy[id][rt]==-1) return;
    lazy[id][rt<<1] = lazy[id][rt<<1|1] = lazy[id][rt];
    if (lazy[id][rt])
    {
        int m = (l+r)>>1;
        tree[id][rt<<1] = m-l+1;
        tree[id][rt<<1|1] = r-m;
    }else tree[id][rt<<1] = tree[id][rt<<1|1] = 0 ;
    lazy[id][rt] = -1;
}
void build(int l,int r,int rt)
{
    if (l==r)
    {
        int id = st[l-1]-'a';
        tree[id][rt] = 1;
        lazy[id][rt] = 1;
        return;
    }
    int m = (l+r)>>1;
    build(lson);
    build(rson);
    for ( int i = 0 ; i < 26 ; i++)
    PushUp(rt,i);
}
 
 
void update(int sc,int l,int r,int rt,int id)
{
    if (r<L||l>R) return;
    if (L<=l&&r<=R)
    {
        lazy[id][rt] = sc;
        if (sc) tree[id][rt] =  r - l + 1;
        else tree[id][rt] = 0;
        return;
    }
    int m = (l+r)>>1;
    PushDown(l,r,rt,id);
    update(sc,lson,id);
    update(sc,rson,id);
    PushUp(rt,id);
}
 
 
 
/*
void update(int L,int R,int sc,int l,int r,int rt,int id)
{
    if (L<=l&&r<=R)
    {
        lazy[id][rt] = sc;
        if (sc) tree[id][rt] = r-l+1;
        else tree[id][rt] =  0;
        return;
    }
    PushDown(l,r,rt,id);
    int m = (l+r)>>1;
    if (L<=m) update(L,R,sc,lson,id);
    if (R>=m+1) update(L,R,sc,rson,id);
    PushUp(rt,id);
}
*/
 
int query(int l,int r,int rt,int id)
{
    if (r<L||l>R) return 0;
    if (L<=l&&r<=R) return tree[id][rt];
    int m = (l+r)>>1;
    PushDown(l,r,rt,id);
    return query(lson,id) + query(rson,id);
}
 
/*
intquery(int L,int R,int l,int r,int rt,int id)
{
    if (L<=l&&r<=R) return tree[id][rt];
    PushDown(l,r,rt,id);
    int m = (l+r)>>1;
    int ret = 0 ;
    if (L<=m) ret = query(L,R,lson,id);
    if (R>=m+1) ret = ret + query(L,R,rson,id);
    return ret;
}
*/
int main()
{
  //     freopen("code/in.txt","r",stdin);
        freopen("input.txt","r",stdin);
        freopen("output.txt","w",stdout);
        ms(tree,0);
        ms(lazy,-1);
        cin>>n>>m;
        cin>>st;
        build(1,n,1);
        while (m--)
        {
            int l,r;
           // scanf("%d%d",&l,&r);
            scan_d(l);scan_d(r);
           // cin>>l>>r;
            int odd = 0;
            int even = 0;
            int InMid=-1;
            for ( int i = 0 ; i < 26 ; i++)
            {
                L = l;
                R = r;
                cnt[i] = query(1,n,1,i);
                if (cnt[i]&1==1) odd++,InMid = i;
                else even++;
                if (odd>1) break;
            }
            int len = r-l+1;
            if (len%2==0)
            {
                if (odd) continue;
            }
            else
            {
                if (odd>1) continue;
            }
            if (odd==1&&even==0) continue;
            //先判断能否回文，再打上重置标记。
            for ( int i = 0 ; i < 26 ; i++)
            {
                L = l;
                R = r;
                update(0,1,n,1,i);
            }
            int fpos = l;
            int bpos = r;
            L = (l+r)/2;
            R = (l+r)/2;
            if (len%2==1) update(1,1,n,1,InMid);
            for ( int i = 0 ; i < 26 ; i++)
            {
                if (cnt[i]<=1) continue;
                L = fpos;
                R = fpos + (cnt[i]>>1)-1;
//                if (L<l||R>(l+r)/2) return 0;
                update(1,1,n,1,i);
                L = bpos - (cnt[i]>>1) + 1;
                R = bpos;
//                if (L<=(l+r)/2||R>r) return 0;
                update(1,1,n,1,i);
                fpos +=cnt[i]>>1;
                bpos-=cnt[i]>>1;
            }
        }
        for ( int i = 1 ; i <= n ; i++)
            for ( int j = 0 ; j < 26 ; j++)
            {
                L = i;
                R = i;
                if (query(1,n,1,j))
                {
                    printf("%c",j+'a');
                 //   cout<<char(j+'a');
                    break;
                }
            }
  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}
 
 
 
 
 

 

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