【拓扑排序】Genealogical tree
[POJ2367]Genealogical tree
Time Limit: 1000MS | Memory Limit: 65536K | |||
Total Submissions: 5696 | Accepted: 3729 | Special Judge |
Description
And in the Planetary Council the confusing genealogical system leads to some embarrassment. There meet the worthiest of Martians, and therefore in order to offend nobody in all of the discussions it is used first to give the floor to the old Martians, than to the younger ones and only than to the most young childless assessors. However, the maintenance of this order really is not a trivial task. Not always Martian knows all of his parents (and there's nothing to tell about his grandparents!). But if by a mistake first speak a grandson and only than his young appearing great-grandfather, this is a real scandal.
Your task is to write a program, which would define once and for all, an order that would guarantee that every member of the Council takes the floor earlier than each of his descendants.
Input
Output
Sample Input
5
0
4 5 1 0
1 0
5 3 0
3 0
Sample Output
2 4 5 3 1
Source
直接拓扑……
代码
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
//#include<cmath> using namespace std;
const int INF = 9999999;
#define LL long long inline int read(){
int x=0,f=1;char c=getchar();
for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
for(;isdigit(c);c=getchar()) x=x*10+c-'0';
return x*f;
}
int N,M;
bool dis[501][501];
bool vis[501];int ru[501]; int main(){
//freopen(".in","r",stdin);
//freopen(".out","w",stdout);
N=read();
for(int i=1;i<=N;i++){
int a=read();
while(a!=0) ru[a]++,dis[i][a]=true,a=read();
}
int tmp;
for(int i=1;i<=N;i++){
if(!ru[i]) {printf("%d ",i);tmp=i;break;}
}
vis[tmp]=true;
for(int i=1;i<N;i++){
for(int j=1;j<=N;j++)
if(dis[tmp][j]) dis[tmp][j]=false,ru[j]--;
for(int j=1;j<=N;j++) if(!ru[j]&&!vis[j]){
vis[j]=true; tmp=j;
if(i!=N-1)printf("%d ",j);
else printf("%d\n",j);
break;
}
}
return 0;
}
【拓扑排序】Genealogical tree的更多相关文章
- 拓扑排序 POJ2367Genealogical tree[topo-sort]
---恢复内容开始--- Genealogical tree Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4875 A ...
- timus 1022 Genealogical Tree(拓扑排序)
Genealogical Tree Time limit: 1.0 secondMemory limit: 64 MB Background The system of Martians’ blood ...
- POJ 2367:Genealogical tree(拓扑排序模板)
Genealogical tree Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7285 Accepted: 4704 ...
- poj 2367 Genealogical tree【拓扑排序输出可行解】
Genealogical tree Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 3674 Accepted: 2445 ...
- POJ 2367:Genealogical tree(拓扑排序)
Genealogical tree Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 2738 Accepted: 1838 Spe ...
- poj 2367 Genealogical tree (拓扑排序)
火星人的血缘关系很奇怪,一个人可以有很多父亲,当然一个人也可以有很多孩子.有些时候分不清辈分会产生一些尴尬.所以写个程序来让n个人排序,长辈排在晚辈前面. 输入:N 代表n个人 1~n 接下来n行 第 ...
- POJ2367 Genealogical tree (拓扑排序)
裸拓扑排序. 拓扑排序 用一个队列实现,先把入度为0的点放入队列.然后考虑不断在图中删除队列中的点,每次删除一个点会产生一些新的入度为0的点.把这些点插入队列. 注意:有向无环图 g[] : g[i] ...
- POJ 2367 Genealogical tree 拓扑排序入门题
Genealogical tree Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8003 Accepted: 5184 ...
- POJ 2367 Genealogical tree【拓扑排序/记录路径】
Genealogical tree Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7101 Accepted: 4585 Spe ...
随机推荐
- 原生ES-Module在浏览器中的尝试
其实浏览器原生模块相关的支持也已经出了一两年了(我第一次知道这个事情实在2016年下半年的时候) 可以抛开webpack直接使用import之类的语法 但因为算是一个比较新的东西,所以现在基本只能自己 ...
- 项目记录 -- python调用回调函数
C源文件: static int get_callback(zpool_handle_t *zhp, void *data) { zprop_get_cbdata_t *cbp = (zprop_ge ...
- Java线程(一)
1. java什么叫线程安全?什么叫不安全? 就是线程同步的意思,就是当一个程序对一个线程安全的方法或者语句进行访问的时候,其他的不能再对他进行操作了,必须等到这次访问结束以后才能对这个线程安全的方法 ...
- Optimizing subroutine calls based on architecture level of called subroutine
A technique is provided for generating stubs. A processing circuit receives a call to a called funct ...
- 爬虫===登陆CSDN的方法
本文主要介绍csdn的登陆,可应用在爬虫上~ # -*- coding:utf-8 -*- import json import requestsfrom xlutils.copy import co ...
- Nginx集群配置与redis的session共享策略
一.什么是Nginx? Nginx (engine x) 是一个高性能的HTTP和反向代理服务器,也是一个IMAP/POP3/SMTP服务器.Nginx是由伊戈尔·赛索耶夫为俄罗斯访问量第二的Ramb ...
- linux服务器上如何显示工作路径
1. 修改PS环境变量 [root@linux-node01 ~]# vi /etc/bashrc [ "$PS1" = "\\s-\\v\\\$ " ] &a ...
- LeetCode218. The Skyline Problem
https://leetcode.com/problems/the-skyline-problem/description/ A city's skyline is the outer contour ...
- numpy及scipy的使用
numpy的使用 把list A转换为numpy 矩阵 np.array(A) np.array(A, 'int32') numpy加载txt文件里面的矩阵 matrix = np.loadtxt(t ...
- php-cgi segmentation fault nginx
谷歌.百度了一堆后,无果. yum安装软件也报segmentation fault 果断重装系统吧