[POJ2367]Genealogical tree

Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 5696   Accepted: 3729   Special Judge

Description

The system of Martians' blood relations is confusing enough. Actually, Martians bud when they want and where they want. They gather together in different groups, so that a Martian can have one parent as well as ten. Nobody will be surprised by a hundred of children. Martians have got used to this and their style of life seems to them natural. 
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

The first line of the standard input contains an only number N, 1 <= N <= 100 — a number of members of the Martian Planetary Council. According to the centuries-old tradition members of the Council are enumerated with the natural numbers from 1 up to N. Further, there are exactly N lines, moreover, the I-th line contains a list of I-th member's children. The list of children is a sequence of serial numbers of children in a arbitrary order separated by spaces. The list of children may be empty. The list (even if it is empty) ends with 0.

Output

The standard output should contain in its only line a sequence of speakers' numbers, separated by spaces. If several sequences satisfy the conditions of the problem, you are to write to the standard output any of them. At least one such sequence always exists.

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;
}

最新文章

  1. linux文件描述符--转载
  2. 使用SDWebImage 怎么获取指定请求对应的缓存图片呢?
  3. sql中truncate 、delete与drop区别
  4. 转:LoadRunner响应时间与用户体验时间不一致问题的深入分析
  5. windows Vista-Server 2016系列激活脚本.cmd
  6. 三菱Ethernet工业以太网
  7. CSS Core Technology
  8. coderwarrior 查看程序大小 Code Size
  9. 2017-12-15python全栈9期第二天第七节之布尔值转数字
  10. 利用Aspectj实现Oval的自动参数校验
  11. 安卓开发_数据存储技术_SharedPreferences类
  12. spring使用@Autowired为抽象父类注入依赖
  13. NIO读写文件并加锁
  14. 解决libc.so.6: version `GLIBC_2.14&amp;#39; not found问题
  15. Express模版引擎hbs备忘
  16. 【BZOJ1434】[ZJOI2009]染色游戏(博弈论)
  17. springboot-web进阶(一)——表单验证
  18. Sysmon + NXlog构建简单的windows安全监控
  19. MySQL学习笔记:case when
  20. python opencv3 grabcut前景检测

热门文章

  1. CRB and Candies(组合数学+求逆元+lcm)
  2. 移动端页面使用rem布局
  3. Python 语言使用中遇到的问题汇总
  4. RecycleView Bug:java.lang.IndexOutOfBoundsException: Inconsistency detected.
  5. web服务器和数据库服务器不在一台机器上
  6. python基础===一道小学奥数题的解法
  7. PXC 避免加入集群时发生SST
  8. python爬虫模块之数据存储模块
  9. npm更新包
  10. centos 时间日期设置