PAT 甲级 1032 Sharing (25 分)(结构体模拟链表,结构体的赋值是深拷贝)

1032 Sharing (25 分)

 

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 (≤), 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 −.

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

题意：

求两个链表的首个共同结点的地址。如果没有，就输出-1

错误点：

1.第一次测试点3超时，原因可能是在遍历时，多次出现a[i].v，a[i].nxt，a[i]出现的次数多了那么遍历a的次数也多了，可能会超时。

2.结构体赋值时深度拷贝，要注意！！！深拷贝是将对象及值复制过来，两个对象修改其中任意的值另一个值不会改变。

#include<bits/stdc++.h>
using namespace std;
struct node{
    int v;
}a[100005];
int main()
{
    a[1].v=520;
    node t=a[1];
    t.v=1314;
    cout<<"a[1].v "<<a[1].v<<endl; 
    cout<<"t.v "<<t.v<<endl; 
    return 0;
}

结果:
a[1].v 520
t.v 1314

AC代码：

#include<bits/stdc++.h>
using namespace std;
struct node{
    char k;
    int nxt;
    int v;
}a[100005];
int main()
{
    int head1,head2,n;
    cin>>head1>>head2>>n;
    int x,y;char s;
    for(int i=1;i<=n;i++){    
        cin>>x>>s>>y;
        a[x].k=s;
        a[x].nxt=y;
        a[x].v=0;
    }
    int f=-1;
    for(int i=head1;i!=-1;i=a[i].nxt){
        a[i].v=1;
    }
    for(int i=head2;i!=-1;i=a[i].nxt){
        if(a[i].v==1){
            f=i;
            break;
        }
    }
    if(f==-1){
        cout<<f;
    }else{
        printf("%05d",f);
    }
    return 0;
}