11029: 【原1029】整理书架

题目

题目描述

author: yydcool 原OJ链接：https://acm.sjtu.edu.cn/OnlineJudge-old/problem/1029

Description

二哥又要整理书架了。他整理书架的方法很简单，把书架上一排所有的书全部挪到另一排的后面。现在二哥把它整理的顺序告诉你，你来告诉他整理之后的书架是什么样子的。

Input Format

读入一个数 \( n \leq 100 \)，表示书架一共有n排，接下来有n行，每行有一些数字(不多于100个数)，每个数字代表一本书，每一行表述这一排书架上的书。再下来有n-1个数对，数对x，y表示把第x排的书放到第y排的后面。

Output Format

输出只有一行，输出最后一排的所有书。

Sample Input

3
1 2 3
4 5
6 7
3 1
2 1

Sample Output

1 2 3 6 7 4 5

FineArtz's solution

/* 整理书架 */
#include <iostream>
#include <sstream>
#include <vector>
using namespace std;

int main(){
    int n;
    cin >> n;
    vector<vector<int>> book;
    string ss;
    getline(cin, ss);
    for (int i = 1; i <= n ; ++i){
        string st;
        getline(cin, st);
        istringstream iss(st);
        int t;
        vector<int> tv;
        while (iss >> t)
            tv.push_back(t);
        book.push_back(tv);
    }
    for (int i = 1; i < n; ++i){
        int x, y;
        cin >> x >> y;
        book[y - 1].insert(book[y - 1].end(), book[x - 1].begin(), book[x - 1].end());
        book[x - 1].clear();
    }
    for (int i = n - 1; i >= 0; --i){
        if (!book[i].empty()){
            for (auto j : book[i]){
                cout << j << ' ';
            }
            cout << endl;
            return 0;
        }
    }
    return 0;
}

ligongzzz's solution

#include "iostream"
#include "cstdio"
#include "cstring"
using namespace std;

class node {
public:
    int val = 0;
    node* next = nullptr;
};

node* heads[109] = { nullptr };
node* rears[109] = { nullptr };

char ch[10009] = { 0 };

int main() {
    int n;
    scanf("%d", &n);

    for (int i = 1; i <= n; ++i) {
        rears[i] = heads[i] = new node;
    }

    cin.getline(ch, 10009);

    for (int i = 1; i <= n; ++i) {
        cin.getline(ch, 10000);
        int length = strlen(ch);

        int num = 0;

        for (int j = 0; j < length; ++j) {
            if (ch[j] == ' ') {
                continue;
            }
            else if (ch[j] == '\n') {
                break;
            }
            else {
                num = num * 10 + ch[j] - '0';
                if (ch[j + 1]<'0' || ch[j + 1]>'9') {
                    rears[i]->next = new node;
                    rears[i] = rears[i]->next;
                    rears[i]->val = num;
                    num = 0;
                }
            }
        }
    }

    for (int i = 0; i < n - 1; ++i) {
        int a, b;
        scanf("%d %d", &a, &b);

        rears[b]->next = heads[a]->next;
        heads[a]->next = nullptr;
        rears[b] = rears[a];
        rears[a] = heads[a];
    }

    for (int i = n; i > 0; --i) {
        if (heads[i]->next == nullptr)
            continue;
        for (auto p = heads[i]->next; p; p = p->next)
            cout << p->val << " ";
    }

    return 0;
}

yyong119's solution

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
int n, i, h, t, indexx;
int main() {
    queue<string> que[101];
    string temp;
    scanf("%d\n", &n);
    for (i = 1; i <= n; i++) {
        getline(cin, temp);
        que[i].push(temp);
        }
    for (i = 1; i < n; i++) {
        scanf("%d%d", &t, &h);
        while (que[t].size()) {
            que[h].push(que[t].front());
            que[t].pop();
        }
    }
    for (indexx = 1; indexx <= n; indexx++) if (que[indexx].size()) break;
    while (que[indexx].size()) {
        cout<<que[indexx].front()<<" ";
        que[indexx].pop();
    }
    return 0;
}