静态链表

上一篇文章说的是动态链表：点击打开链接

动态链表需要用指针来建立结点之间的联系，如果结点的地址是比较小的整数的话，就没有必要去建立动态链表，而应该使用更方便的静态链表。静态链表的实现原理是hash，即通过一个建立一个结构体数组，并令数组的下标直接表示结点的地址，来达到直接访问数组中的元素就能访问结点的效果。对于静态链表，由于结点的访问时非常方便的，所以不需要头结点。静态链表的定义方法如下：

struct Node{typename data; //数据域 int next;  //指针域 }node[size];

例题：

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.

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 Nextis 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 967890 i 0000200010 a 1234500003 g -112345 D 6789000002 n 0000322222 B 2345611111 L 0000123456 e 6789000001 o 00010

Sample Output 1:

67890

Sample Input 2:

00001 00002 400001 a 1000110001 s -100002 a 1000210002 t -1

Sample Output 2:

-1

题解：

#include <cstdio>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;}