題目連結：http://www.patest.cn/contests/pat-a-practise/1032

題目：

1032. Sharing (25)

時間限制 100 ms

記憶體限制 65536 kB

代碼長度限制 16000 B

判題程式 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}, andNext 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      

分析：

兩個連結清單。找到它們相交的第一個節點。

注意：

可能存在多個字串。而且可能有其它非題目中兩字串的節點存在。而且注意最後是要輸出%05d的

AC代碼：

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int nodes[100001];
int real_nodes[100001];
int main(void){
 //freopen("F://Temp/input.txt", "r", stdin);
 int s1, s2, n;
 scanf("%d %d %d", &s1, &s2, &n);
 int s, t;
 char c;
 for (int i = 0; i < 100001; i++){
  nodes[i] = 0;
  real_nodes[i] = 0;
 }
 for (int i = 0; i < n; i++){
  scanf("%d %c %d", &s, &c, &t);
  nodes[s] = t;
 }
 int start = s1; int end;
 while (start != -1){
  real_nodes[start] = 1;
  start = nodes[start];
 }//找到第一個連結清單的全部節點，并置為1
 start = s2;
 bool find = false;
 while (start != -1){
  if (real_nodes[start] != 0){//假設第二個連結清單中位址已經有記錄，則就是第一個公共節點
   printf("%.5d\n", start);
   return 0;
  }
  else {
   real_nodes[start] = nodes[start];
   start = nodes[start];
  }
 }
 puts("-1");
 return 0;
}
           

截圖：

——Apie陳小旭