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 Nextis 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
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 using namespace std; 5 typedef struct nd{ 6 int rt; 7 char alfa; 8 int isFirst; 9 nd(){ 10 isFirst = 0; 11 } 12 }node; 13 node data[100000]; 14 int main(){ 15 int w1, w2, N; 16 scanf("%d%d%d", &w1, &w2, &N); 17 int temp1, temp2; 18 char c; 19 for(int i = 0; i < N; i++){ 20 scanf("%d %c %d", &temp1, &c, &temp2); 21 data[temp1].alfa = c; 22 data[temp1].rt = temp2; 23 } 24 int pt = w1; 25 while(pt != -1){ 26 data[pt].isFirst = 1; 27 pt = data[pt].rt; 28 } 29 int find = -1; 30 pt = w2; 31 while(pt != -1){ 32 int tag = 1, pt2 = pt; 33 if(data[pt].isFirst == 1){ 34 while(pt2 != -1){ 35 if(data[pt2].isFirst != 1){ 36 tag = 0; 37 break; 38 } 39 pt2 = data[pt2].rt; 40 } 41 if(tag == 1){ 42 find = pt; 43 break; 44 } 45 } 46 pt = data[pt].rt; 47 } 48 if(find == -1) 49 printf("-1"); 50 else printf("%05d", find); 51 cin >> N; 52 return 0; 53 }
总结：
1、在地址很小的时候，为了方便一般使用静态数组。