Codeforces Codeforces Round #316 (Div. 2) C. Replacement 线段树
C. Replacement
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/570/problem/C
Description
Daniel has a string s, consisting of lowercase English letters and period signs (characters '.'). Let's define the operation of replacementas the following sequence of steps: find a substring ".." (two consecutive periods) in string s, of all occurrences of the substring let's choose the first one, and replace this substring with string ".". In other words, during the replacement operation, the first two consecutive periods are replaced by one. If string s contains no two consecutive periods, then nothing happens.
Let's define f(s) as the minimum number of operations of replacement to perform, so that the string does not have any two consecutive periods left.
You need to process m queries, the i-th results in that the character at position xi (1 ≤ xi ≤ n) of string s is assigned value ci. After each operation you have to calculate and output the value of f(s).
Help Daniel to process all queries.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 300 000) the length of the string and the number of queries.
The second line contains string s, consisting of n lowercase English letters and period signs.
The following m lines contain the descriptions of queries. The i-th line contains integer xi and ci (1 ≤ xi ≤ n, ci — a lowercas English letter or a period sign), describing the query of assigning symbol ci to position xi.
Output
Print m numbers, one per line, the i-th of these numbers must be equal to the value of f(s) after performing the i-th assignment.
Sample Input
10 3
.b..bz....
1 h
3 c
9 f
Sample Output
4
3
1
HINT
题意
给你一个字符串,然后每两个点可以变成一个点
然后有m次修改操作,每次可以修改一个位置的字符
然后问你修改之后,需要多少次操作,把所有的点,都变成连续的一个点
题解:
我比较蠢,我用的线段树,太蠢了
正解不超过30行,几个if就好了……
维护的是每一个pos的左边的字母和右边的字母位置
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
using namespace std; #define LL(x) (x<<1)
#define RR(x) (x<<1|1)
#define MID(a,b) (a+((b-a)>>1))
const int N=; struct node
{
int lft,rht;
int lmx,rmx;
int len(){return rht-lft+;}
int mid(){return MID(lft,rht);}
void init(){lmx=rmx=len();}
void fun(int valu)
{
if(valu==-) lmx=rmx=;
else lmx=rmx=;
}
}; int n,m; struct Segtree
{
node tree[N*];
void up(int ind)
{
tree[ind].lmx=tree[LL(ind)].lmx;
tree[ind].rmx=tree[RR(ind)].rmx;
if(tree[LL(ind)].lmx==tree[LL(ind)].len())
tree[ind].lmx+=tree[RR(ind)].lmx;
if(tree[RR(ind)].rmx==tree[RR(ind)].len())
tree[ind].rmx+=tree[LL(ind)].rmx;
}
void build(int lft,int rht,int ind)
{
tree[ind].lft=lft,tree[ind].rht=rht;
tree[ind].init();
if(lft!=rht)
{
int mid=tree[ind].mid();
build(lft,mid,LL(ind));
build(mid+,rht,RR(ind));
}
}
void updata(int pos,int ind,int valu)
{
if(tree[ind].lft==tree[ind].rht) tree[ind].fun(valu);
else
{
int mid=tree[ind].mid();
if(pos<=mid) updata(pos,LL(ind),valu);
else updata(pos,RR(ind),valu);
up(ind);
}
}
void query(int pos,int ind,int& x,int& y)
{
if(tree[ind].lft==tree[ind].rht)
{
if(tree[ind].lmx==) x=y=tree[ind].lft;
else x=y=;
}
else
{
int mid=tree[ind].mid();
if(pos<=mid) query(pos,LL(ind),x,y);
else query(pos,RR(ind),x,y);
if(tree[LL(ind)].rht==y) y+=tree[RR(ind)].lmx;
if(tree[RR(ind)].lft==x) x-=tree[LL(ind)].rmx;
}
}
}seg;
char s[N];
int vis[N];
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
scanf("%s",s+);
seg.build(,n+,);
int len=strlen(s+);
int ans=;
int tmp=;
seg.updata(n+,,-);
for(int i=;i<=len;i++)
{
if(s[i]=='.')
tmp++;
else
{
if(tmp!=)
ans+=(tmp-);
tmp=;
seg.updata(i,,-);
vis[i]=;
}
}
if(tmp!=)
ans+=(tmp-);
while(m--)
{
char cmd[];
int pos,st,ed;
scanf("%d",&pos);
scanf("%s",&cmd);
if(cmd[]!='.')
{
if(vis[pos]==)
printf("%d\n",ans);
else
{
vis[pos]=; seg.query(pos,,st,ed);
ans-=(ed-st);
seg.updata(pos,,-);
seg.query(pos+,,st,ed);
ans+=(ed-st);
if(pos-!=)
{
seg.query(pos-,,st,ed);
ans+=(ed-st);
}
printf("%d\n",ans);
} }
else
{
if(vis[pos]==)
printf("%d\n",ans);
else
{
vis[pos]=;
seg.query(pos+,,st,ed);
ans-=(ed-st);
if(pos-!=)
{
seg.query(pos-,,st,ed);
ans-=(ed-st);
}
seg.updata(pos,,);
seg.query(pos,,st,ed);
ans+=(ed-st);
printf("%d\n",ans); }
}
}
}
return ;
}
Codeforces Codeforces Round #316 (Div. 2) C. Replacement 线段树的更多相关文章
- Codeforces Round #603 (Div. 2) E. Editor 线段树
E. Editor The development of a text editor is a hard problem. You need to implement an extra module ...
- Codeforces Round #406 (Div. 1) B. Legacy 线段树建图跑最短路
B. Legacy 题目连接: http://codeforces.com/contest/786/problem/B Description Rick and his co-workers have ...
- Codeforces Round #765 Div.1 F. Souvenirs 线段树
题目链接:http://codeforces.com/contest/765/problem/F 题意概述: 给出一个序列,若干组询问,问给出下标区间中两数作差的最小绝对值. 分析: 这个题揭示着数据 ...
- 【转】Codeforces Round #406 (Div. 1) B. Legacy 线段树建图&&最短路
B. Legacy 题目连接: http://codeforces.com/contest/786/problem/B Description Rick and his co-workers have ...
- Codeforces Codeforces Round #316 (Div. 2) C. Replacement set
C. Replacement Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/570/proble ...
- Codeforces Round #316 (Div. 2) C. Replacement
题意:给定一个字符串,里面有各种小写字母和' . ' ,无论是什么字母,都是一样的,假设遇到' . . ' ,就要合并成一个' .',有m个询问,每次都在字符串某个位置上将原来的字符改成题目给的字符, ...
- Codeforces Round #316 (Div. 2) C. Replacement(线段树)
C. Replacement time limit per test 2 seconds memory limit per test 256 megabytes input standard inpu ...
- Codeforces Round #316 (Div. 2C) 570C Replacement
题目:Click here 题意:看一下题目下面的Note就会明白的. 分析:一开始想的麻烦了,用了树状数组(第一次用)优化,可惜没用. 直接判断: #include <bits/stdc++. ...
- Codeforces Round #316 (Div. 2) C Replacement 扫描法
先扫描一遍得到每个位置向后连续的'.'的长度,包含自身,然后在扫一遍求出初始的合并次数. 对于询问,只要对应位置判断一下是不是'.',以及周围的情况. #include<bits/stdc++. ...
随机推荐
- 【转】使用 Auto Layout 的典型痛点和技巧
layoutIfNeeded()强制立刻更新布局 原文网址:http://www.jianshu.com/p/0f031606e5f2 官方文档:Auto Layout Guide 加上去年WWDC上 ...
- Ruby基础数据类型
#数字分为证书Integer,浮点数Float(对应与其他语言中的double),和复数Complex #整数又分为Fixnum和Bignum,Fixnum和Bignum会互相转换,这些都是ruby自 ...
- JavaScript的事件代理(转)
如果你想给网页添加点JavaScript的交互性,也许你已经听过JavaScript的事件代理(event delegation),并且觉得这是那些发烧友级别的JavaScript程序员才会关心的什么 ...
- Java SE 6 新特性: Java DB 和 JDBC 4.0
http://www.ibm.com/developerworks/cn/java/j-lo-jse65/index.html 长久以来,由于大量(甚至几乎所有)的 Java 应用都依赖于数据库,如何 ...
- Loadrunner模拟Json请求
一.loadrunner脚本创建 1.Insert - New step -选择Custom Request - web_custom_request 2.填入相应参数 3.生成脚本,并修改如下(参数 ...
- DIV+CSS常见问题:DIV如何设置一个像素高度?
CSS如何控制DIV实现1像素高度呢?问题看起来很简单,但万恶的IE6会让你很麻烦,不过解决办法很多,本文将介绍最简单的一种:DIV{height:1px;line-height:1px;font-s ...
- 在DataTable 中增加一列
//在这里需要增加一个列. DataColumn column = dt.Columns.Add("行号", Type.GetType("S ...
- aspx与ascx,ashx的用法详细的总结介绍
这篇文章主要是对aspx与ascx,ashx的用法进行了详细的总结介绍,需要的朋友可以过来参考下,希望对大家有所帮助 做asp.net开发的对.aspx,.ascx和.ashx都不会陌生.关于它们,网 ...
- 机器学习----分布问题(二元,多元变量分布,Beta,Dir)
这涉及到数学的概率问题. 二元变量分布: 伯努利分布,就是0-1分布(比如一次抛硬币,正面朝上概率) 那么一次抛硬币的概率分布如下: 假设训练数据如下: 那么根据最大似然估计(MLE ...
- RabbitMQ>Erlang machine stopped instantly (distribution name conflict?). The service is not restarted as OnFail is set to ignore.-报错解决方案 原来是NNND。。。
>Erlang machine stopped instantly (distribution name conflict?). The service is not restarted as ...