PAT 甲级 1032 Sharing (25分)

题目

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 (≤10⁵), 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

给出两个单词的字母的起始地址和字母个数N，接下来N行为：地址 字母 下一个字母的地址；
求出两个单词共同后缀开始的字母的地址，没有则输出-1；

思路

这种链表题可以使用数组去分别存储地址中的元素，和该地址的下一个字母的地址，先将第一个单词的所有地址存到set，然后对另一个单词从起始地址开始遍历直到找到相同的地址，没有则输出-1；

代码

#include<bits/stdc++.h>
using namespace std;
int main(int argc, const char * argv[]) {
    int st1, st2, N, next[100010], a, b, pos = -1;
    char ch;
    scanf("%d%d%d", &st1, &st2, &N);
    for(int i=0; i<N; i++){
        cin>>a>>ch>>b;
        next[a] = b;
    }
    set<int> nextAddrs;
    int tmp = st2;
    while(tmp != -1){
        nextAddrs.insert(tmp);
        tmp = next[tmp];
    }
    tmp = st1;
    while(tmp != -1){
        if(nextAddrs.count(tmp)){
            pos = tmp;
            break;
        }
        tmp = next[tmp];
    }
    if(pos == -1) printf("-1");
    else printf("%05d", pos);
    return 0;
}