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
- 查找yum 安装目录
[root@DB-Server ~]# rpm -qa | grep -i mysql MySQL-server-advanced-5.6.20-1.rhel5 MySQL-client-advanc ...
- Spring MVC的handlermapping之请求分发如何找到正确的Handler(BeanNameUrlHandlerMapping,SimpleUrlHandlerMapping)
本文讲的是Spring MVC如何找到正确的handler, 前面请求具体怎么进入到下面的方法,不再细说. 大概就是Spring mvc通过servlet拦截请求,实现doService方法,然后进入 ...
- sklearn中决策树算法DesiciontTreeClassifier()调用以及sklearn自带的数据包sklearn.datasets.load_iris()的应用
决策树方法的简单调用记录一下 clf=tree.DecisionTreeClassifier() dataMat=[];labelMat=[] dataPath='D:/machinelearning ...
- DevExpress,LayoutControl,TreeList,GridControl等
1.显示边框进行折叠 选择一个layoutControlGroupX 将其GroupBordersVisible设置成True,将TextVisiable=True 2. TreeList 2.1需要 ...
- Kali linux切换语言为中文
echo LANG="zh_CN.UTF-8" > /etc/default/locale
- Linux 基础教程 27-ss和ip命令
什么是netstat 在Linux系统中输入 man netstat,显示的结果如下所示: netstat - Print network connections, routing table ...
- 使用VMWare12.0安装Ubuntu系统
使用VMWare12.0安装Ubuntu系统 Vmware12的虚拟机的文档说明: http://pubs.vmware.com/workstation-12/index.jsp#com.vmware ...
- sonar扫描多模块的maven项目
SonarQube Scanner,作为代码扫描的工具,通过它,将项目的代码读取并发送至SonarQube服务器中,才能让SonarQube进行代码分析. 可以认为SonarQube Scanner就 ...
- .Net MVC5路由机制与扩展
新建一个MVC项目启动后,首先访问的地址是http://localhost:xxx/Home/Index,这时候我们也明白因为在程序中有个叫做Home的控制器,并且在这个控制器下面有个叫做Index的 ...
- 纸壳CMS列表Grid的配置
纸壳CMS(ZKEACMS)里的Grid是一个TagHelper,是对jQuery插件datatables的一个配置封装. Easy.Mvc.TagHelpers.GridTagHelper grid ...