1032. Sharing (25)-优快云博客

本文介绍了一种算法，用于解决两个由节点组成的英语单词链表如何找到它们共享后缀的开始位置的问题。该算法通过一次遍历来标记每个节点，并再次遍历来确定共享后缀的起始地址。

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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<string>
using namespace std;
int main(){
	int S1,S2;
	int N;
	cin>>S1>>S2>>N;
	int i;
	int a[100000]={0};
	int b[100000]={-1};
	int shuffer=-1;
	int temp1,temp3;
	char temp2;
	for(i=0;i<N;i++){
		scanf("%d %c %d",&temp1,&temp2,&temp3);
		if((temp2>='a'&&temp2<='z')||(temp2>='A'&&temp2<='Z'))
		b[temp1]=temp3;
	}
	while(S1!=-1){
		a[S1]++;
		S1=b[S1];
		//cout<<S1<<" ";
	}
	
	while(S2!=-1){
		a[S2]++;
		if(a[S2]==2){
		shuffer=S2;
		break;
	}
		S2=b[S2];
		//cout<<S2<<" ";
	}
	if(shuffer!=-1)
	printf("%05d",shuffer);
	else
	cout<<-1;
	}

感想：老老实实模拟，测试点自有坑你的地方，注意不能用cout会超时  然后注意处理链表只有一个元素的情况