Sharing-链表

题目描述

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).

输入描述:

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
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.

输出描述:

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.

示例1

输入

复制

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

输出

复制

67890
-1

//链表的定义，用链表做
#include<iostream>
#include<vector>
#include<string.h>
#include<stdio.h>
using namespace std;
struct node
{
    char c;
    int next;//指针指向的下一个地址
    int flag;//记录该节点被访问的次数,将访问次数变成2的结点的地址return，否则输出-1
};
vector<node>lists;//存放结点，vector的下标就是地址，有一点hash的感觉
int  getindex(vector<node>&lists,int start1,int start2)
{
    int i;
    i=start1;
    while(i!=-1)
    {
        lists[i].flag++;
        i=lists[i].next;
    }
    i=start2;
     while(i!=-1)
    {
        lists[i].flag++;
        if(lists[i].flag==2)
        {
            return i;
        }
        i=lists[i].next;
    }
     return -1;
}
int main()
{
    int start1,start2,N;
    while(cin>>start1>>start2>>N)
    {
        int address;
        lists.clear();
        lists.resize(100000);//增大vector容量，必须的！！！！否则的话会提示段错误。当结点的个数超过256，都resize一下吧！
        for(int i=0;i<N;i++)
        {
            lists[i].flag=0;
        }
        for(int i=0;i<N;i++)
        {
            cin>>address;
            cin>>lists[address].c>>lists[address].next;

        }
        int index=getindex(lists,start1,start2);
        if(index!=-1)
        {
            printf("%05d\n",index);
        }
        else cout<<index<<endl;
    }

}