A1032 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.

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 -1instead.

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：

给出两条链表的首地址以及若干个节点的地址、数据、下一个节点的地址，求两条链表的首个共用节点的地址。如果两条链表没有共用节点，则输出 -1. N (≤10​^5​​)

思路：

由于地址范围很小，可以使用静态数组。

定义 Node 结构体中定义 flag ，表示节点是否在第一条链表中出现过，若出现，赋值为1

枚举第二条链表，若出现第一个 flag 值为1的节点，即为共用节点；如果枚举完仍然没有发现共用节点，输出 -1

#include <cstdio>
#include <cstring>
const int maxn = 100010;
struct Node{
    char data;
    int next;
    bool flag;
}node[maxn];
int main(){
    for(int i = 0; i < maxn; i++){
        node[i].flag = false;
    }
    int s1, s2, n;
    scanf("%d%d%d", &s1, &s2, &n);
    int address, next;
    char data;
    for(int i = 0; i < n; i++){
        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){
        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;
}