1032 Sharing (输出链表的公共结点的地址)

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.

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 (≤10^​5), 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

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:

Sample Input 2:

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

Sample Output 2:

题意：

给出两个链表的首地址和结点个数以及若干结点的地址、数据、下一个结点的地址，求两个链表首个共用结点的地址，如果没有共用结点，输出-1。

思路：

由于地址的范围很小，可以考虑静态链表。依题意，应在结点的结构体中多加一个bool变量flag表示结点是否在第一个链表中出现（初始化flag=0），如果出现flag=1，否则flag=0。

注意：

scanf使用%c格式时会读入空格，所以在输入地址、数据及后继结点地址时，格式不能写成%d%c%d，必须在中间加空格，即%d %c %d

#include <cstdio>

const int maxn = 100010;
struct NODE
{
    char data;		//数据
    int  next;		//下一个结点的地址
    bool flag;
}node[maxn];

int main()
{
    for(int i = 0; i < maxn; i++)	//将所有结点flag初始化为false
        node[i].flag = false;
    int s1, s2, n;
    scanf("%d%d%d", &s1, &s2, &n);	//s1，s2分别是两个链表的首地址，n是结点个数
    int address, next;
    char data;
    for(int i = 0 ; i < n; i++)		//输入结点信息
    {
        scanf("%d %c %d", &address, &data, &next);
        node[address].data = data;
        node[address].next = next;
    }
    int p;
    for(p = s1; p != -1; p = node[p].next)	//遍历第一个链表
        node[p].flag = true;
    for(p = s2; p != -1; p = node[p].next)
        if(node[p].flag == true)
            break;
    if(p != -1)
        printf("%05d\n", p);
    else
        printf("-1\n");
    return 0;
}