Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem C: Close Relatives

Roger is researching the family tree for his pet quesnot Arthur. Quesnots are mythical animals who cooperate to raise a number of young. Adult quesnots form a parenthood of up to ten parents to raise a litter of young. Once the young have matured, all parents are exhausted and cannot help raise a second litter before they die.

Roger has determined some of Arthur's ancestors' families, but is having trouble determining whether certain relatives are ancestors of another. He has written a program to calculate this fact, but Roger requires lists of a particular format. Your job is to convert his family tree into one or more lists for him to use.

Input

A family tree will be given as input. Each line will consist of a quesnot name, possibly followed by a list of at most ten parents. Each name will be at most 30 characters, and may include letters, numbers, hyphens, spaces or periods. Names on a line will be separated by commas. There will be at least one and at most 5000 unique names in the tree, and no quesnot can be his own ancestor.

Recall that each quesnot will have at most one child. Brothers and sisters, aunts and uncles will not be included in the input. Each name will occur as the first name on a line at most once. You may assume that the first name on the first line is the bottom of the tree (i.e., will have no children), and that this is the only name in the file with no children.

Output

You are to produce a minimum number of lists with the following properties:

• Every quesnot mentioned in the tree occurs in each list exactly once,
• Quesnot A is an ancestor of quesnot B exactly when A appears before B in every list.

The first line of output indicates the number of lists m, m > 0. A blank line then appears. Following are m lists, each containing all names in appropriate order. A blank line appears between lists.

There may be several solutions satisfying these requirements. Any such solution may be chosen.

Sample Input

Arthur,Bob,Julie
Bob,Adam,Jane
Jane,Monica,George
Julie,Pat

Output for Sample Input

2

Adam
Monica
George
Jane
Bob
Pat
Julie
Arthur

Pat
Julie
George
Monica
Jane
Adam
Bob
Arthur

---

Richard Krueger

題目描述：

當給定一個樹狀關係，當長輩都死亡後，該人才可死亡。

給定最少種情況的死亡順序，可以推論出原本的樹狀關係。

題目解法：

一種是優先左探訪，另一種優先右探訪。

即可推論出長輩關係

#include <stdio.h>
#include <string.h>
#include <map>
#include <vector>
#include <iostream>
using namespace std;
vector<int> g[5005];
string D[5005];
void dfs1(int nd) {
    int i;
    for(i = 0; i < g[nd].size(); i++)
        dfs1(g[nd][i]);
    printf("%s\n", D[nd].c_str());
}
void dfs2(int nd) {
    int i;
    for(i = g[nd].size()-1; i >= 0; i--)
        dfs2(g[nd][i]);
    printf("%s\n", D[nd].c_str());
}
int main() {
    char s[5005];
    int i, j, k;
    int size = 0;
    map<string, int> R;
    while(gets(s)) {
        string x, y;
        int len = strlen(s);
        for(i = 0; s[i] != ',' && s[i]; i++);
        s[i] = '\0', x = s;
        int &vx = R[x];
        if(vx == 0) vx = ++size;
        while(i < len) {
            for(j = ++i; s[i] != ',' && s[i]; i++);
            s[i] = '\0';
            y = s+j;
            int &vy = R[y];
            if(vy == 0) vy = ++size;
            g[vx].push_back(vy);
        }
    }
    for(map<string, int>::iterator it = R.begin();
        it != R.end(); it++)
            D[it->second] = it->first;
    puts("2\n");
    dfs1(1);
    puts("");
    dfs2(1);
    return 0;
}