Mouse Hunt CodeForces - 1027D(思维 找环)
Medicine faculty of Berland State University has just finished their admission campaign. As usual, about 80%80% of applicants are girls and majority of them are going to live in the university dormitory for the next 44 (hopefully) years.
The dormitory consists of nn rooms and a single mouse! Girls decided to set mouse traps in some rooms to get rid of the horrible monster. Setting a trap in room number ii costs cici burles. Rooms are numbered from 11 to nn.
Mouse doesn't sit in place all the time, it constantly runs. If it is in room ii in second tt then it will run to room aiai in second t+1t+1 without visiting any other rooms inbetween (i=aii=ai means that mouse won't leave room ii). It's second 00 in the start. If the mouse is in some room with a mouse trap in it, then the mouse get caught into this trap.
That would have been so easy if the girls actually knew where the mouse at. Unfortunately, that's not the case, mouse can be in any room from 11 to nn at second 00.
What it the minimal total amount of burles girls can spend to set the traps in order to guarantee that the mouse will eventually be caught no matter the room it started from?
Input
The first line contains as single integers nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of rooms in the dormitory.
The second line contains nn integers c1,c2,…,cnc1,c2,…,cn (1≤ci≤1041≤ci≤104) — cici is the cost of setting the trap in room number ii.
The third line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤n1≤ai≤n) — aiai is the room the mouse will run to the next second after being in room ii.
Output
Print a single integer — the minimal total amount of burles girls can spend to set the traps in order to guarantee that the mouse will eventually be caught no matter the room it started from.
Examples
5
1 2 3 2 10
1 3 4 3 3
3
4
1 10 2 10
2 4 2 2
10
7
1 1 1 1 1 1 1
2 2 2 3 6 7 6
2
Note
In the first example it is enough to set mouse trap in rooms 11 and 44. If mouse starts in room 11 then it gets caught immideately. If mouse starts in any other room then it eventually comes to room 44.
In the second example it is enough to set mouse trap in room 22. If mouse starts in room 22 then it gets caught immideately. If mouse starts in any other room then it runs to room 22 in second 11.
Here are the paths of the mouse from different starts from the third example:
- 1→2→2→…1→2→2→…;
- 2→2→…2→2→…;
- 3→2→2→…3→2→2→…;
- 4→3→2→2→…4→3→2→2→…;
- 5→6→7→6→…5→6→7→6→…;
- 6→7→6→…6→7→6→…;
- 7→6→7→…7→6→7→…;
So it's enough to set traps in rooms 22 and 66.
一个连通块 肯定存在一个环
所以找环上的最小值即可
为什么一个连通块肯定存在一个环 因为每个点都有一个出度 。。。
#include <bits/stdc++.h>
#define mem(a, b) memset(a, b, sizeof(a))
using namespace std;
const int maxn = 1e5 + , INF = 0x7fffffff;
typedef long long LL;
int n, cnt;
int a[maxn<<], head[maxn<<], vis[maxn<<], pre[maxn<<];
int s, t;
struct node
{
int u, v, next;
}Node[maxn<<]; void add(int u, int v)
{
Node[cnt].u = u;
Node[cnt].v = v;
Node[cnt].next = head[u];
head[u] = cnt++;
} void dfs1(int u, int fa)
{
vis[u] = ;
for(int i=head[u]; i!=-; i=Node[i].next)
{
node e = Node[i];
if(!vis[e.v])
{
pre[e.v] = u;
dfs1(e.v, u);
}
else
{
s = e.v;
t = u;
return;
}
}
}
int minn = INF;
LL res = ;
void dfs2(int u) //由t向e.v回溯,如果能回溯到s则说明这是一个新的环 那么就把res += minn 其实放到一个dfs里就好了
{
minn = min(minn, a[u]); if(u == s)
{
res += (LL)minn;
return;
}
else if(u == ) return;
dfs2(pre[u]);
} int main()
{
mem(head, -);
cnt = ;
cin>> n;
for(int i=; i<=n; i++)
cin>> a[i];
int u;
for(int i=; i<=n; i++)
{
cin>> u;
add(i, u);
}
for(int i=; i<=n; i++)
{
if(!vis[i])
{
minn = INF;
d[i] = ;
dfs1(i, -);
dfs2(t); }
}
cout<< res <<endl; return ;
}
Mouse Hunt CodeForces - 1027D(思维 找环)的更多相关文章
- 【CF1027D】Mouse Hunt(拓扑排序,环)
题意:给定n个房间,有一只老鼠可能从其中的任意一个出现, 在第i个房间设置捕鼠夹的代价是a[i],若老鼠当前在i号房间则下一秒会移动到b[i]号, 问一定能抓住老鼠的最小的总代价 n<=2e5, ...
- 【Edu49 1027D】 Mouse Hunt DFS 环
1027D. Mouse Hunt:http://codeforces.com/contest/1027/problem/D 题意: 有n个房间,每个房间放置捕鼠器的费用是不同的,已知老鼠在一个房间x ...
- Codeforces 1027D Mouse Hunt (强连通缩点 || DFS+并查集)
<题目链接> 题目大意: 有n个房间,每个房间都会有一只老鼠.处于第i个房间的老鼠可以逃窜到第ai个房间中.现在要清理掉所有的老鼠,而在第i个房间中防止老鼠夹的花费是ci,问你消灭掉所有老 ...
- Codeforces B. Mouse Hunt(强连通分解缩点)
题目描述: Mouse Hunt time limit per test 2 seconds memory limit per test 256 megabytes input standard in ...
- Codeforces Beta Round #88 C. Cycle —— DFS(找环)
题目链接:http://codeforces.com/problemset/problem/117/C C. Cycle time limit per test 2.5 seconds memory ...
- Codeforces Round #369 (Div. 2) D. Directed Roads —— DFS找环 + 快速幂
题目链接:http://codeforces.com/problemset/problem/711/D D. Directed Roads time limit per test 2 seconds ...
- 【CodeForces】915 D. Almost Acyclic Graph 拓扑排序找环
[题目]D. Almost Acyclic Graph [题意]给定n个点的有向图(无重边),问能否删除一条边使得全图无环.n<=500,m<=10^5. [算法]拓扑排序 [题解]找到一 ...
- CF1027D Mouse Hunt 思维
Mouse Hunt time limit per test 2 seconds memory limit per test 256 megabytes input standard input ou ...
- CodeForces 711D Directed Roads (DFS找环+组合数)
<题目链接> 题目大意: 给定一个$n$条边,$n$个点的图,每个点只有一条出边(初始状态),现在能够任意对图上的边进行翻转,问你能够使得该有向图不出先环的方案数有多少种. 解题分析: 很 ...
随机推荐
- C51中的关键字和ANSIC标准关键字
C51中的关键字和ANSIC标准关键字 作 者:武力戡乱 修改日期:2017-09-05 备 注: 1.总备注信息 2.联系方式 3.其它博文链接:武力戡乱博客目录总表 内 ...
- 手机视频编辑软件APP
1. VUE iOS/Android 2.Alive iOS/Android 3.Splice iOS 4. Plotagragh+ 能让照片动起来的app 5.Cinepic 能 ...
- 处女男学Android(七)---Android 应用资源之StateListDrawable
前言 本篇Blog将记录关于Android应用资源中最经常使用的一个Drawable资源--StateListDrawable,本来说应当继续写UI方面的内容,突然跳到应用资源这边,主要是由于之前写界 ...
- WC 2019 游记 - 败者之低语
败者之低语 WC 2019 游记 Day -1 看了一圈PKU和THU的题,感觉图像识别真有意思... 感觉非常讲道理,pku还是比thu简单一点的... 听说高二414在thu没有进面试? 震惊!( ...
- Python3入门(十)——调试与测试
一.异常处理 1.try...except...finally... 这个也就是Java里的try...cath..finally...了,直接看经典代码: try: print("开始执行 ...
- 大数据入门第十二天——sqoop入门
一.概述 1.sqoop是什么 从其官网:http://sqoop.apache.org/ Apache Sqoop(TM) is a tool designed for efficiently tr ...
- 20155204《网络对抗》Exp7 网络欺诈防范
20155204<网络对抗>Exp7 网络欺诈防范 一.基础问题回答 1.通常在什么场景下容易受到DNS spoof攻击 在不安全的网络环境下访问网站. 2.在日常生活工作中如何防范以上两 ...
- 20155302 Exp2 后门原理与实践
20155302<网络对抗>后门原理与实践 实验要求 1.使用netcat获取主机操作Shell,cron启动 (0.5分) 2.使用socat获取主机操作Shell, 任务计划启动 (0 ...
- # 2017-2018-2 20155319『网络对抗技术』Exp6:信息收集与漏洞扫描
2017-2018-2 20155319『网络对抗技术』Exp6:信息收集与漏洞扫描 实践内容 (1)各种搜索技巧的应用 (2)DNS IP注册信息的查询 (3)基本的扫描技术:主机发现.端口扫描.O ...
- 20155334 《网络攻防》Exp5 MSF基础应用
一.基础问题回答 解释exploit,payload,encode是什么: 项目 作用 exploit 是负载有用代码的交通工具,让代码到达目的地,并作用 payload 是有具体功能的代码,能够完成 ...