zoukankan      html  css  js  c++  java
  • PAT Advanced 1032 Sharing (25分)

    To store English words, one method is to use linked lists and store a word letter by letter. To save some space, we may let the words share the same sublist if they share the same suffix. For example, loading and being are stored as showed in Figure 1.

    fig.jpg

    Figure 1

    You are supposed to find the starting position of the common suffix (e.g. the position of i in Figure 1).

    Input Specification:

    Each input file contains one test case. For each case, the first line contains two addresses of nodes and a positive N (≤), where the two addresses are the addresses of the first nodes of the two words, and N is the total number of nodes. The address of a node is a 5-digit positive integer, and NULL is represented by −.

    Then N lines follow, each describes a node in the format:

    Address Data Next
    
     

    whereAddress is the position of the node, Data is the letter contained by this node which is an English letter chosen from { a-z, A-Z }, and Next is the position of the next node.

    Output Specification:

    For each case, simply output the 5-digit starting position of the common suffix. If the two words have no common suffix, output -1 instead.

    Sample Input 1:

    11111 22222 9
    67890 i 00002
    00010 a 12345
    00003 g -1
    12345 D 67890
    00002 n 00003
    22222 B 23456
    11111 L 00001
    23456 e 67890
    00001 o 00010
    
     

    Sample Output 1:

    67890
    
     

    Sample Input 2:

    00001 00002 4
    00001 a 10001
    10001 s -1
    00002 a 10002
    10002 t -1
    
     

    Sample Output 2:

    -1
    最后一个case错是因为,没有理解,同一个字母可以有多个地址
    #include <iostream>
    #include <unordered_map>
    using namespace std;
    struct node{
        string addr, ch, next;
    };
    int main() {
        ios::sync_with_stdio(false);
        cin.tie(0);
        int N;
        string addr, addr2;
        node tmp;
        unordered_map<string, node> m;
        unordered_map<string, bool> m2;
        cin >> addr >> addr2 >> N;
        for(int i = 0; i < N; i++) {
            cin >> tmp.addr >> tmp.ch >> tmp.next;
            m[tmp.addr] = tmp;
        }
        while(addr != "-1") {
            m2[addr] = true;
            addr = m[addr].next;
        }
        while(addr2 != "-1") {
            if(m2[addr2]) {
                cout << m[addr2].addr << endl;return 0;
            }
            addr2 = m[addr2].next;
        }
        cout << -1 << endl;return 0;
    }
  • 相关阅读:
    VS中的 MD/MT设置 【转】
    VS2010/MFC编程入门之五十四(Ribbon界面开发:使用更多控件并为控件添加消息处理函数)
    VS2010/MFC编程入门之五十三(Ribbon界面开发:为Ribbon Bar添加控件)[转]
    [MFC]选择目录对话框和选择文件对话框 [转]
    NMM3DViewer 设计
    将可执行程序的内存空间扩展到3GB(windows)
    centos7 安装rocketmq(quick start)
    Centos7 安装 Maven 3.5.*
    ss命令
    强制重启Linux系统的几种方法
  • 原文地址:https://www.cnblogs.com/littlepage/p/12263723.html
Copyright © 2011-2022 走看看