1032 Sharing (25分)
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).
Each input file contains one test case. For each case, the first line contains two addresses of nodes and a positive N (≤), 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 −.
Then N lines follow, each describes a node in the format:
Address Data Next
whereAddress 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.
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.
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
67890
#include<cstdio> #include<iostream> using namespace std; const int N = 100000; struct node { char data; int next; bool flag; }node[N]; int main() { for (int i = 0; i < N; i++) { node[i].flag = false; } int s1, s2, n; scanf("%d %d %d",&s1,&s2,&n); int address, next; char data; for (int j = 0; j < n; j++) { scanf("%d %c %d", &address, &data, &next); node[address].data = data; node[address].next = next; } int p; for (p = s1; p != -1; p = node[p].next) { node[p].flag = true; } for (p = s2; p != -1; p = node[p].next) { if (node[p].flag == true) break; } if (p != -1) printf("%05d\n", p); else printf("-1\n"); return 0; }
思路2
不同部分为a， 和b，公共部分为c；a + c + b = b + c + a;让两个一起走，a走到头就转向b， b走到头转向a，则在公共部分相遇。
对于没有公共节点 则在尾结点处停止
#include<cstdio> #include<iostream> using namespace std; const int N = 100000; struct node { char data; int next; bool flag; }node[N]; int main() { for (int i = 0; i < N; i++) { node[i].flag = false; } int s1, s2, n; scanf("%d %d %d",&s1,&s2,&n); int address, next; char data; for (int j = 0; j < n; j++) { scanf("%d %c %d", &address, &data, &next); node[address].data = data; node[address].next = next; } int p=s1,q=s2; while (p != q) { if (p != -1) p = node[p].next; else p = s2; if (q != -1) q = node[q].next; else q = s1; } printf("%d\n", p); }
原文：https://www.cnblogs.com/lifutao/p/12881008.html