poj2367 Genealogical tree(拓扑排序:输出方案)
2016-02-02 15:49
351 查看
思路:模板题
一时SB忘记加!=EOF,一直OLE
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
Sample Output
一时SB忘记加!=EOF,一直OLE
#include <cstdio> #include <queue> #include <cstring> #include <iostream> #include <cstdlib> #include <algorithm> #include <vector> #include <map> #include <string> #include <set> #include <ctime> #include <cmath> #include <cctype> using namespace std; #define maxn 1000 #define LL long long int cas=1,T; int n; vector<int>G[maxn]; int in[maxn]; int ans[maxn]; void topo() { queue<int>q; for (int i = 1;i<=n;i++) if (in[i]==0) q.push(i); int cnt=0; while (!q.empty()) { int u = q.front(); q.pop(); ans[cnt++]=u; for (int i = 0;i<G[u].size();i++) { int v = G[u][i]; if (--in[v]==0) q.push(v); } } } int main() { while (scanf("%d",&n)!=EOF && n) { for (int i = 1;i<=n;i++) { G[i].clear(); in[i]=0; } for (int i = 1;i<=n;i++) { int v; while (scanf("%d",&v) && v) { G[i].push_back(v); in[v]++; //记得入度要加一 } } topo(); printf("%d",ans[0]); for (int i = 1;i<n;i++) printf(" %d",ans[i]); // puts(""); printf("\n"); } //freopen("in","r",stdin); //scanf("%d",&T); //printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC); return 0; }
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
相关文章推荐
- 自动 更新SVN目录文件.bat
- 安装MySQL Workbench 6.3.4 CE (winx64)出错KERNELBASE.dll的解决方案
- Canvas
- htmlcleaner使用及xpath语法
- Quartz1.8.5例子(八)
- Java的图片处理工具类
- java接口和抽象类的区别和作用(功能、用途、好处)
- jquery怎样循环获得一组id值
- 新浪微盘3
- 编辑pdf内容
- html转成pdf的方法
- 我的IT人生-2015
- android 编译ffmpeg
- 人的核心竞争力超过一半来自重要而不紧急的事情:
- 绝对定位让元素居中
- Win8.1系统把时间设置为12小时制的方法
- 11155 ly与lyon的终极巅峰对决
- 【慕课笔记】U2 封装 第8节 JAVA中的方法内部类
- Android5.0 style样式新特性
- 算法题13 排序算法(更新快排)