hihoCoder1343 : Stable Members【BFS拓扑排序】
题目链接:https://hihocoder.com/problemset/problem/1343
#1343 : Stable Members
描述
Recently Little Hi joined an algorithm learning group. The group consists of one algorithm master and N members. The members are numbered from 1 to N. Each member has one or more other members as his mentors. Some members' mentor is the master himself.
Every week each member sends a report of his own learning progress and the reports collected from his pupils (if there is any) to his mentors. The group is so well designed that there is no loop in the reporting chain so no one receives his own report from his pupil. And finally the master gets every one's report (maybe more than once).
Little Hi notices that for some members their reporting routes to the master can be easily cut off by a single member's (other than the master and himself) absence from the reporting duty. They are called unstable members while the others are stable members. Given the reporting network of the group, can you find out how many members are stable?
Assume there are 4 members in the group. Member 1 and 2 both have the master as their only mentor. Member 3 has 2 mentors: member 1 and member 2. Member 4 has 1 mentor: member 3. Then member 4 is the only unstable member in the group because if member 3 is absent his learning report will be unable to be sent to the master.
输入
The first line contains an integer N, the number of members.
The i-th line of the following N lines describe the mentors of the i-th member. The first integer is Ki, the number of mentors of the i-th member. Then follows Ki integers A1 ... AN, which are his mentors' numbers. Number 0 indicates that the master is one of his mentor.
For 40% of the data, 1 ≤ N ≤ 1000.
For 100% of the data, 1 ≤ N ≤ 100000.
For 100% of the data, 1 ≤ Ki ≤ 10, Ki < N, 0 ≤ Ai ≤ N.
输出
Output the number of stable members.
- 样例输入
-
5
1 0
1 0
2 1 2
1 3
2 4 3 - 样例输出
-
3

#include<bits/stdc++.h>
using namespace std;
const int N = ;
struct node {
int color = ;
vector<int>s, p;//子节点、父节点
}a[N];
bool unstable[N];
bool all_colored(int v, int color) {
int num = a[v].p.size();
bool flag = true;
for(int i = ; flag && i < num; ++i)
flag &= (a[a[v].p[i]].color == color);
return flag;
}
void topo(int v) {
if(unstable[v]) return;
queue<int>q;
q.push(v);
a[v].color = v;
while(!q.empty()) {
int u = q.front(); q.pop();
int num = a[u].s.size();
for(int i = ; i < num; ++i) {
int son = a[u].s[i];
if(all_colored(son, v)) {
a[son].color = v;
unstable[son] = true;
q.push(son);
}
}
}
}
int main() {
int n, k, i, v, ans = ;
scanf("%d", &n);
for(i = ; i <= n; ++i) {
scanf("%d", &k);
while(k--) {
scanf("%d", &v);
a[i].p.push_back(v);
a[v].s.push_back(i);
}
}
for(i = ; i <= n; ++i) topo(i);
for(i = ; i <= n; ++i) ans += unstable[i];
printf("%d\n", n - ans);
return ;
}
hihoCoder1343 : Stable Members【BFS拓扑排序】的更多相关文章
- hihocoder 1343 : Stable Members【拓扑排序】
hihocoder #1343:题目 解释:一个学习小组,一共有N个学员,一个主管.每个学员都有自己的导师(一个或者多个),导师可以是其他学员也可以是主管.每周学员都要把自己的学习报告和收到的报告提交 ...
- C. Journey bfs 拓扑排序+dp
C. Journey 补今天早训 这个是一个dp,开始我以为是一个图论,然后就写了一个dij和网络流,然后mle了,不过我觉得如果空间开的足够的,应该也是可以过的. 然后看了题解说是一个dp,这个dp ...
- uvaLA4255 Guess BFS+拓扑排序
算法指南白书 思路:“连续和转化成前缀和之差” #include <stdio.h> #include <string.h> #include <iostream> ...
- CH 2101 - 可达性统计 - [BFS拓扑排序+bitset状压]
题目链接:传送门 描述 给定一张N个点M条边的有向无环图,分别统计从每个点出发能够到达的点的数量.N,M≤30000. 输入格式 第一行两个整数N,M,接下来M行每行两个整数x,y,表示从x到y的一条 ...
- hihocoder 1174 [BFS /拓扑排序判断是否有环]
hihocoder 1174 [算法]: 计算每一个点的入度值deg[i],这一步需要扫描所有点和边,复杂度O(N+M). 把入度为0的点加入队列Q中,当然有可能存在多个入度为0的点,同时它们之间也不 ...
- Going from u to v or from v to u?_POJ2762强连通+并查集缩点+拓扑排序
Going from u to v or from v to u? Time Limit: 2000MS Memory Limit: 65536K Description I ...
- 【ZOJ - 3780】 Paint the Grid Again (拓扑排序)
Leo has a grid with N × N cells. He wants to paint each cell with a specific color (either black or ...
- BFS (1)算法模板 看是否需要分层 (2)拓扑排序——检测编译时的循环依赖 制定有依赖关系的任务的执行顺序 djkstra无非是将bfs模板中的deque修改为heapq
BFS模板,记住这5个: (1)针对树的BFS 1.1 无需分层遍历 from collections import deque def levelOrderTree(root): if not ro ...
- hdu1532 用BFS求拓扑排序
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1285 题目给出一些点对之间的先后顺序,要求给出一个字典序最小的拓扑排列.对于拓扑排序的问题,我们有DF ...
随机推荐
- iOS 中的 armv7,armv7s,arm64,i386,x86_64 都是什么
在做静态库的时候以及引用静态库的时候经常会遇到一些关于真机模拟器不通用的情况,会报错找不到相应库导致编译失败, 这里简单记录一下各种设备支持的架构. iOS测试分为模拟器测试和真机测试,处理器分为32 ...
- 啰里吧嗦jvm
一.为什么要了解jvm 有次做项目的时候,程序run起来的时候,总是报OutOfMemoryError,有老司机教我们用jconsole.exe看内存溢出问题 就是这货启动jconsole后,发现一个 ...
- 【转】到底什么时候应该用MQ
原文地址:http://zhuanlan.51cto.com/art/201704/536407.htm 一.缘起 一切脱离业务的架构设计与新技术引入都是耍流氓. 引入一个技术之前,首先应该解答的问题 ...
- Java反射拾遗
定义:Java反射机制可以让我们在编译期(Compile Time)之外的运行期(Runtime)检查类,接口,变量以及方法的信息.反射还可以让我们在运行期实例化对象,调用方法,通过调用get/set ...
- C# 设计模式·行为型模式
这里列举行为型模式·到此23种就列完了···这里是看着菜鸟教程来实现··,他里边列了25种,其中过滤器模式和空对象模式应该不属于所谓的23种模式责任链模式:为请求创建一个接收者对象的链,对请求的发送者 ...
- cakephp中使用大括号的形式避免用点号连接sql语句
在cakephp中可以使用{}的形式来代替点号连接sql语句,减少出错的几率
- 计时器(Chronometer)
计时器(Chronometer) 常用属性:format(计时器的计时格式) 常用方法: setBase(long base) 设置计时器的起始时间 setFormat(String format) ...
- dev gridview指定单元格cell获取坐标
DevExpress.XtraGrid.Views.Grid.ViewInfo.GridViewInfo Info2 = gvQueryResult.GetViewInfo() as DevExpre ...
- 更新oracle数据库表如何实现主键自增长
在数据库中实现主键自动增长有利于我们做数据插入操作,在SQL SERVER上创建表时可以在int类型的字段后加上identity(1,1),该字段就会从1开始,按照+1的方式自增,将这个字段设置 ...
- android 常见分辨率(mdpi、hdpi 、xhdpi、xxhdpi )及屏幕适配
1 Android手机目前常见的分辨率 1.1 手机常见分辨率: 4:3VGA 640*480 (Video Graphics Array)QVGA 320*240 (Quarter VGA ...