原题链接:A1032 Sharing

貌似是某年408数据结构算法考研题。(解析如下:)
PAT (Advanced Level) Practice A1032 Sharing (25 分)(C++)(甲级)(链表存储相同后缀的单词)(考研题)

#include<algorithm>
#include<iostream>
#include<cstdio>
using namespace std;

typedef struct Link
{
    char data;
    int next;
}Link;
Link L[100010];//L以地址存放

int main()
{
    int F1, F2, N, ad, ad1, ad2, len1 = 0, len2 = 0, i;
    scanf("%d %d %d", &F1, &F2, &N);
    for(i=0; i<N; i++)
    {
        scanf("%d", &ad);
        getchar();//小心空格被吸收掉
        scanf("%c %d", &L[ad].data, &L[ad].next);
    }
    for(ad = F1; ad != -1; ad = L[ad].next, len1++);//计算长度
    for(ad = F2; ad != -1; ad = L[ad].next, len2++);
    for(ad1 = F1, i = 0; i < len1-len2; ad1 = L[ad1].next, i++);//让两者从倒着数相同的位置开始
    for(ad2 = F2, i = 0; i < len2-len1; ad2 = L[ad2].next, i++);
    while(ad1 != ad2)
    {
        ad1 = L[ad1].next;
        ad2 = L[ad2].next;
    }
    if(ad1 == -1)printf("-1");
    else printf("%05d", ad1);//还是要考虑格式,最后一个测试用例是这个坑……
    return 0;
}

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