CodeForces - 556C-Case of Matryoshkas(思维)
Andrewid the Android is a galaxy-famous detective. He is now investigating the case of vandalism at the exhibition of contemporary art.
The main exhibit is a construction of n matryoshka dolls that can be nested one into another. The matryoshka dolls are numbered from 1 to n. A matryoshka with a smaller number can be nested in a matryoshka with a higher number, two matryoshkas can not be directly nested in the same doll, but there may be chain nestings, for example, 1 → 2 → 4 → 5.
In one second, you can perform one of the two following operations:
- Having a matryoshka a that isn't nested in any other matryoshka and a matryoshka b, such that b doesn't contain any other matryoshka and is not nested in any other matryoshka, you may put a in b;
- Having a matryoshka a directly contained in matryoshka b, such that b is not nested in any other matryoshka, you may get a out of b.
According to the modern aesthetic norms the matryoshka dolls on display were assembled in a specific configuration, i.e. as several separate chains of nested matryoshkas, but the criminal, following the mysterious plan, took out all the dolls and assembled them into a single large chain (1 → 2 → ... → n). In order to continue the investigation Andrewid needs to know in what minimum time it is possible to perform this action.
Input
The first line contains integers n (1 ≤ n ≤ 105) and k (1 ≤ k ≤ 105) — the number of matryoshkas and matryoshka chains in the initial configuration.
The next k lines contain the descriptions of the chains: the i-th line first contains number mi (1 ≤ mi ≤ n), and then mi numbers ai1, ai2, ..., aimi — the numbers of matryoshkas in the chain (matryoshka ai1 is nested into matryoshka ai2, that is nested into matryoshka ai3, and so on till the matryoshka aimi that isn't nested into any other matryoshka).
It is guaranteed that m1 + m2 + ... + mk = n, the numbers of matryoshkas in all the chains are distinct, in each chain the numbers of matryoshkas follow in the ascending order.
Output
In the single line print the minimum number of seconds needed to assemble one large chain from the initial configuration.
Examples
Input
3 2
2 1 2
1 3
Output
1
Input
7 3
3 1 3 7
2 2 5
2 4 6
Output
10
Note
In the first sample test there are two chains: 1 → 2 and 3. In one second you can nest the first chain into the second one and get 1 → 2 → 3.
In the second sample test you need to disassemble all the three chains into individual matryoshkas in 2 + 1 + 1 = 4 seconds and then assemble one big chain in 6 seconds.
思路:链上从1开始连续的数字不用拆。一旦遇到不连续,则从这个不连续的娃娃开始后面的全部都要拆。
代码:
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<cmath>
const int maxn=1e5+5;
typedef long long ll;
using namespace std;
int a[maxn];
int main()
{
int n,k;
cin>>n>>k;
int m;
ll sum1=0,sum2=0;
for(int t=0;t<k;t++)
{
scanf("%d",&m);
int s=1;
int x,x1;
scanf("%d",&a[0]);
int flag=0;
if(a[0]==1)
{
flag=1;
}
for(int k=1;k<m;k++)
{
scanf("%d",&a[k]);
if(flag==1&&(a[k]-a[k-1]==1))
{
continue;
}
else
{
flag=0;
s++;
sum2++;
}
}
sum1+=s;
}
sum1=sum1-1+sum2;
if(sum1<0)
{
cout<<0<<endl;
}
else
cout<<sum1<<endl;
return 0;
}
CodeForces - 556C-Case of Matryoshkas(思维)的更多相关文章
- CodeForces - 556C Case of Matryoshkas
//////////////////////////////////////////////////////////////////////////////////////////////////// ...
- CodeForces - 556C Case of Matryoshkas (水题)
Andrewid the Android is a galaxy-famous detective. He is now investigating the case of vandalism at ...
- codeforces 556C. Case of Matryoshkas 解题报告
题目链接:http://codeforces.com/contest/556/problem/C 题目意思:有 n 个数(1,2,...,n)组成 k 条链.第 i 条链由 mi 个数组成.每一秒只可 ...
- codeforces 555a//Case of Matryoshkas// Codeforces Round #310(Div. 1)
题意:1-n的数字,大的在小的后面,以这种规则已经形成的几个串,现在要转为一个串,可用的操作是在末尾拆或添加,问要操作几次? 模拟了很久还是失败,看题解才知道是数学.看来这种只要结果的题,模拟很不合算 ...
- 贪心/思维题 Codeforces Round #310 (Div. 2) C. Case of Matryoshkas
题目传送门 /* 题意:套娃娃,可以套一个单独的娃娃,或者把最后面的娃娃取出,最后使得0-1-2-...-(n-1),问最少要几步 贪心/思维题:娃娃的状态:取出+套上(2),套上(1), 已套上(0 ...
- 「日常训练」Case of Matryoshkas(Codeforces Round #310 Div. 2 C)
题意与分析(CodeForces 556C) 为了将所有\(n\)个娃娃编号递增地串在一起(原先是若干个串,每个串是递增的), 我们有两种操作: 拆出当前串中最大编号的娃娃(且一定是最右边的娃娃). ...
- [Codeforces 555E]Case of Computer Network(Tarjan求边-双连通分量+树上差分)
[Codeforces 555E]Case of Computer Network(Tarjan求边-双连通分量+树上差分) 题面 给出一个无向图,以及q条有向路径.问是否存在一种给边定向的方案,使得 ...
- 【35.37%】【codeforces 556C】Case of Matryoshkas
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- Codeforces Round #310 (Div. 1) A. Case of Matryoshkas 水题
C. String Manipulation 1.0 Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contes ...
随机推荐
- radiobutton 选中的项不能去掉选择的问题
代码如下: RadioButton rbtn = new RadioButton(getApplicationContext()); rbtn.setText(String.valueOf(item. ...
- 使用cmd命令登录mysql数据库时报2013-Lost connection to MYSQL server at 'waiting for initial communication packet',system error:0
[错误内容]:SQL Error (2013): Lost connection to MySQL server at 'waiting for initial communication packe ...
- xib下这种方式创建cell
这种方法在iOS5.0之前是不能够创建成功的. MEConvertListTableViewCell *cell = [tableView dequeueReusableCellWithIde ...
- 网络排错与网络命令的理解ping-traceroute-host(nslookup)-tcpdump获取对方的mac
1. 虚拟机中NAT架构的网络结构中, 虚拟网卡VMnet8(192.168.134.1)是连接宿主主机. 用虚拟网段中主机(192.168.134.133),ping VMnet8 为什么没有响 ...
- nginx 配置中的if判断
正则表达式匹配: ==:等值比较; ~:与指定正则表达式模式匹配时返回“真”,判断匹配与否时区分字符大小写: ~*:与指定正则表达式模式匹配时返回“真”,判断匹配与否时不区分字 ...
- 在 Docker 上运行一个 RESTful 风格的微服务
tags: Microservice Restful Docker Author: Andy Ai Weibo:NinetyH GitHub: https://github.com/aiyanbo/d ...
- UTF8和UCS2
谈谈Unicode编码,简要解释UCS.UTF.BMP.BOM等名词 这是一篇程序员写给程序员的趣味读物.所谓趣味是指可以比较轻松地了解一些原来不清楚的概念,增进知识,类似于打RPG游戏的升级.整理这 ...
- 使用 IIS 在 Windows 上托管 ASP.NET Core(Windows安装实践)
原文地址 https://docs.microsoft.com/zh-cn/aspnet/core/host-and-deploy/iis/?view=aspnetcore-2.0&tabs= ...
- pageadmin 网站建设系统如何新建进程池并在站点中使用
1.打开iis管理界面,右键应用程序池,点击添加应用程序池,添加界面如下图,注意pageadmin cms net版本选择4.0,托管模式建议选择集成模式. 2.添加完毕后,在网站中点击对应站点,点击 ...
- Android Studio显示可视化编辑界面
选中项目,依次展开“src/main/res/layout",双击"activity_main.xml",这样右侧就有“preview”选项卡了,点击activity_m ...