1032. Sharing (25)

题目地址：http://www.patest.cn/contests/pat-a-practise/1032
链表考研真题

/*1032. Sharing (25)http://www.patest.cn/contests/pat-a-practise/1032利用deque 很容易实现 不过采用链表LNode的实现差不多也类似，写起来麻烦点*/#include <stdio.h>#include <stdlib.h>#include <math.h>#include <vector>#include <string>#include <unordered_map>#include <queue>#include <algorithm>#include <iostream>using namespace std;#define N 100005int addr1 , addr2 , n ;struct data{    int addr ;    char da ;     int nextaddr;};data tdata[N];int main(){    //freopen("in.txt","r",stdin);    scanf("%d%d%d",&addr1,&addr2,&n);    int i , j ;    int a1,a2;    char c;    for(i = 0 ; i < n ; i ++)    {        scanf("%d %c %d",&a1,&c,&a2);        tdata[a1].addr = a1;        tdata[a1].da = c;        tdata[a1].nextaddr = a2;    }    deque<data> vd1,vd2;    int t = addr1;    while(t != -1)    {        vd1.push_back(tdata[t]);        t = tdata[t].nextaddr;    }    t = addr2;    while(t != -1)    {        vd2.push_back(tdata[t]);        t = tdata[t].nextaddr;    }    int ans = -1 ;    while(!vd1.empty() && !vd2.empty())    {        data d1 = vd1.back() ;        data d2 = vd2.back() ;        if(d1.da == d2.da && d1.addr == d2.addr && d1.nextaddr == d2.nextaddr)        {            ans = d1.addr;            vd1.pop_back();            vd2.pop_back();        }else{            break;        }    }    if(ans == -1)        printf("-1");    else        printf("%05d\n",ans);    return 0;}