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 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
#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);
else
printf("-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*/
