1032. Sharing (25)

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
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

以链表的形式给出两个字符串，其中可能有公共部分，求出公共部分的第一个点的地址。这里跟字母没什么关系，只要用地址就行了。想法就是先建立好链表，然后遍历两个链表找出第一个两个链表都能访问的点。一开始用map做，最后一个case会超时。由于节点地址最多为5位数，所以用数组来储存下一个节点即可，用next表示。然后根据next数组遍历第一个链表，记录哪些点访问过，然后遍历第二个链表，一旦遇到访问过的点就退出循环，输出该点的地址，如果到最后都没遇到访问过的点，即没有公共部分，输出-1.

代码：

#include <iostream>
#include <vector>
#include <cstring>
#include <cstdio>
#include <cstdlib>
using namespace std;

int main()
{
int next[100001];
bool vis[100001];
memset(next,-1,100001);
memset(vis,false,100001);
int a1,a2,n;
scanf("%d %d %d",&a1,&a2,&n);
for (int i=0;i<n;i++)
{
int a,b;
char c;
scanf("%d %c %d",&a,&c,&b);
next[a]=b;
}
int index=a1;
while(index!=-1)
{
vis[index]=true;
index=next[index];
}
index=a2;
while(index!=-1&&!vis[index])
{
index=next[index];
}
if(index==-1)
printf("-1");
else
printf("%05d",index);
}