1032 Sharing (25 分) c++

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 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 (≤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 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
结尾无空行
Sample Output 1:

67890
结尾无空行
Sample Input 2:

00001 00002 4
00001 a 10001
10001 s -1
00002 a 10002
10002 t -1
结尾无空行
Sample Output 2:

-1
结尾无空行

思路：

先遍历第一条链表，之后在遍历第二条链表，将两者进行对比。
注意：对于这种存在对比的题目，我们可以设立一个标识位flag，通过对比遍历前后的flag，来获取信息

具体代码如下：

#include 

#define maxn 100010
struct Node{
    char data;
    int next;
    int flag = -1;
}node[maxn];
int main(){
    int word1, word2, N;
    int add, next;
    char data;
    scanf("%d %d %d", &word1, &word2, &N);
    for(int i = 1; i <= N; i++){
        scanf("%d %c %d", &add, &data, &next);
        node[add].data = data;
        node[add].next = next;
    }
    int p = word1;
    while(p != -1){
        node[p].flag = 1;
        p = node[p].next;
    }
    p = word2;
    while(p != -1){
        if(node[p].flag == 1){
            break;
        }
        p = node[p].next;
    }
    if(p != -1){
        printf("%05dn",p);
    }else{
        printf("-1n");
    }
    
    
    return 0;
}

这里使用了静态链表，之所以不使用动态链表，原因是：对于节点地址比较小的整数（小于五位数），就可以直接使用静态链表～还有输出的数字有位数限制，注意得用%05d来限制输出的位数