1032 Sharing

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.

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

 

题意：

　　用链表来表示两个英文单词，如果两个链表的后缀相同，则可以被两个单词共用。求出相同后缀的第一个单词的Address。

思路：

　　模拟。

Code：

#include <bits/stdc++.h>

using namespace std;

int main() {
    int s1, s2, n;
    cin >> s1 >> s2 >> n;
    map<int, int> m;
    int address, next;
    char data;
    for (int i = 0; i < n; ++i) {
        cin >> address >> data >> next;
        m[address] = next;
    }
    int len1 = 0, len2 = 0;
    int temp1 = s1, temp2 = s2;
    while (temp1 != -1) {
        len1++;
        temp1 = m[temp1];
    }
    while (temp2 != -1) {
        len2++;
        temp2 = m[temp2];
    }
    while (len1 != len2) {
        if (len1 > len2) {
            len1--;
            s1 = m[s1];
        } else {
            len2--;
            s2 = m[s2];
        }
    }
    bool found = false;
    while (len1 > 0) {
        if (s1 == s2) {
            found = true;
            cout << setw(5) << setfill('0') << s1 << endl;
            break;
        } else {
            s1 = m[s1];
            s2 = m[s2];
            len1--;
        }
    }
    if (!found) cout << "-1" << endl;
    return 0;
}

 

最后一组数据刚开始提交的时候超时了，但是提交了几次之后竟然通过了，搞不懂。