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 (≤10e5), 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:

67890

Sample Input 2:

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

Sample Output 2:

-1

题目大意：求两个静态链表的首个共同结点的地址。如果没有，就输出-1。
分析：先遍历第一个链表并标记，再遍历第二个链表，若遍历的节点有标记，则输出该地址；若最后也没碰到标记，则输出-1。

注意第二个链表可能是第一个链表的子链表。

11111 22222 5
11111 b 12345
12345 e 22222
22222 i 00003
00003 n 00004
00004 g -1

#include<algorithm>
#include <iostream>
#include  <cstdlib>
#include  <cstring>
#include   <cstdio>
#include    <ctime>
#include    <cmath>
using namespace std;

typedef struct node
{
    int next,flag;
    char c;
}node;
node num[100000];

int main(void)
{
    #ifdef test
    freopen("in.txt","r",stdin);
    //freopen("in.txt","w",stdout);
    clock_t start=clock();
    #endif //test

    int a,b,n,t,ans=-1;scanf("%d%d%d",&a,&b,&n);

    int l1,l2;
    for(int i=0;i<n;++i)
    {
        int add,next;
        char c;
        scanf("%d %c %d",&add,&c,&next);
        num[add].c=c,num[add].next=next;
    }
    t=a;
    while(t!=-1)
    {
        num[t].flag=1;t=num[t].next;
    }
    t=b;
    while(t!=-1)
    {
        if(num[t].flag==1)
        {
            ans=t;break;
        }
        num[t].flag=1;t=num[t].next;
    }
    if(ans==-1)printf("%d\n",ans);
    else printf("%05d\n",ans);

    #ifdef test
    clockid_t end=clock();
    double endtime=(double)(end-start)/CLOCKS_PER_SEC;
    printf("\n\n\n\n\n");
    cout<<"Total time:"<<endtime<<"s"<<endl;        //s为单位
    cout<<"Total time:"<<endtime*1000<<"ms"<<endl;    //ms为单位
    #endif //test
    return 0;
}