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…

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 store…

1032 Sharing (25分) 题目 思路 定义map存储所有的<地址1,地址2> 第一set存放单词1的所有地址(通过查找map) 通过单词二的首地址,结合map,然后在set中查找是否存在,如果存在就是所求的地址,没有就是-1 注意点 无 代码 #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector> #in…

题目 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 Figu…

题意: 输入两个单词的起始地址和一个正整数N(<=1e5),然后输入N行数据,每行包括一个五位数的字母地址,字母和下一个字母的地址.输出这两个单词的公共后缀首字母的地址,若无公共后缀则输出-1. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; string a1,a2; int n; string s,nex; char x; map<strin…

1032 Sharing (25)(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 "…

1032. Sharing (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue 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 the…

1043 Is It a Binary Search Tree (25 分)   A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a…

题目 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 sufix. For example, "loading" and "being" are stored…

#include <iostream> #include <cstdio> #include <algorithm> #include <string.h> #include <cmath> using namespace std; /* 链表题 水 */ int n; struct Word{ int addr; char ch; ; }word[]; ]; int main() { int first1,first2; int adr,nxt…