1032. Sharing (25)

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本文介绍了一种算法，用于解决两个由节点组成的单词共享相同后缀的问题，并找到该共享后缀的起始位置。输入包括两个单词的首个节点地址及所有节点的数量，每个节点包含地址、字母数据及指向下一个节点的指针。通过遍历节点并比较来找出共同后缀的起始位置。

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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 they share the same suffix. For example, “loading” and “being” are stored as showed in Figure 1.
这里写图片描述
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 (<= 105), 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 -1.
Then N lines follow, each describes a node in the format:
Address Data Next
where Address 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

注意点：

输出地址要进行补零
对静态数组的初始化，一定要为-1

#define _CRT_SECURE_NO_WARNINGS
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <map>
#include <set>

using namespace std;

const int MaxN = 100010;

int main()
{
#ifdef _DEBUG
    freopen("data.txt", "r+", stdin);
#endif // _DEBUG
    int s1, s2, n, addr, next, L1 = 0, L2 = 0, abL;
    int SL[MaxN] = {-1};
    char data;
    std::ios::sync_with_stdio(false);
    cin >> s1 >> s2 >> n;

    for (int i = 0; i < n; ++i)
    {
        cin >> addr >> data >> next;
        SL[addr] = next;
    }

    for (addr = s1; SL[addr] != -1; addr = SL[addr], ++L1);
    for (addr = s2; SL[addr] != -1; addr = SL[addr], ++L2);

    if (L2 > L1)
    {
        swap(L1, L2);
        swap(s1, s2);
    }

    abL = L1 - L2;

    for (int i = 0; i < abL; ++i)
        s1 = SL[s1];

    while (s1 != -1 && s1 != s2)
    {
        s1 = SL[s1];
        s2 = SL[s2];
    }

    if (s1 == -1)
        printf("-1");
    else
        printf("%05d", s1);


    return 0;
}