PAT甲级练习1032. Sharing (25)

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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
一道有关链表的题,一开始想用map来做,但是有超时,就放弃了,之后用了struct来做,对s1访问过的点使用flag标记,之后s2开始判断经过的点flag为true的就是要输出的编号

#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <map>#include <set>#include <string>#include <string.h>using namespace std;struct Node{//Node():data('0'),next(0),flag(false){}char data;int next;bool flag;}node[100000];int main(){int s1, s2, n;scanf("%d %d %d", &s1, &s2, &n);int address, next;char data;for(int i=0; i<n; i++){scanf("%d %c %d", &address, &data, &next);node[address].next = next;node[address].data = data;}while(s1!=-1){node[s1].flag = true;s1 = node[s1].next;}while(s2!=-1){if(node[s2].flag==1){printf("%05d", s2);cin>>n;return 0;}s2 = node[s2].next;}printf("-1\n");cin>>n;return 0;}

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