PAT甲级1032

1032. Sharing (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

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.

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 967890 i 0000200010 a 1234500003 g -112345 D 6789000002 n 0000322222 B 2345611111 L 0000123456 e 6789000001 o 00010

Sample Output 1:

67890

Sample Input 2:

00001 00002 400001 a 1000110001 s -100002 a 1000210002 t -1

Sample Output 2:

-1

#include<iostream>#include<cstdio>#include<queue>#include<vector>#include<algorithm>using namespace std;struct LinkNode{int address, next;char data;};int a[100000];int main(){fill(a, a + 100000, -1);//用于链接int start1, start2, N;cin >> start1 >> start2 >> N;LinkNode ln;//vector<LinkNode> v;for (int i = 0; i < N; i++){cin >> ln.address >> ln.data >> ln.next;//v.push_back(ln);a[ln.address] = ln.next;}int start = start1, len1 = 0, len2 = 0;while (start!=-1){len1++; start = a[start];}start = start2;while (start != -1){len2++; start = a[start];}int difference = abs(len1 - len2);if (len2 > len1){swap(start1, start2);}//以便于之后统一处理，总使得第一个单词长度不短于第二个单词start = start1; int startt = start2;int i = 0;while (start != -1){if (start == startt)break;if (i >=difference){startt = a[startt];}start = a[start];i++;}//cout << start;if (start != -1)printf("%05d", start);elseprintf("-1");return 0;}/*这个就是先找出两个单词的距离之差，然后让长的先移这么多。之后再同步移，并比较地址*/