PAT 1003 Sharing (25)

题目描写叙述

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).

输入描写叙述:

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.

输出描写叙述:

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.

输入样例:

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

输出样例:

67890

//最初的解法低效。且有未知错误 /*思路例如以下： 从最后面開始找。直到存在某个前驱有不止一个后继指向它，则结束*/ #include <iostream> #include <iomanip>using namespace std;typedef struct Node {int addressPre,addressPost;char data; };int main() {int a1,a2,N,i,j,k,ans;while(cin>>a1>>a2>>N){Node *node=new Node[N];for(i=0;i<N;i++)cin>>node[i].addressPre>>node[i].data>>node[i].addressPost; /*for(i=0;i<N;i++)cout<<node[i].addressPre<<" "<<node[i].data<<" "<<node[i].addressPost<<endl; */k=-1;while(1){ans=0;for(i=0;i<N;i++){if(k==node[i].addressPost){ans++;j=i;} }if(ans>1)break;elsek=node[j].addressPre;}cout <<setfill('0')<< setw(5)<<k<<endl;}return 0; }

正解

#include <iostream> #include <iomanip> #include <vector>using namespace std;typedef struct Node {int addressPre,addressPost;char data; };vector<Node> list1; vector<Node> list2;Node node[100010];int main() {int a1,a2,N,i,j,k;int l1,l2;Node tNode;while(cin>>a1>>a2>>N){for(i=0;i<N;i++){cin>>tNode.addressPre>>tNode.data>>tNode.addressPost;node[tNode.addressPre]=tNode;}l1=a1;l2=a2;while(l1!=-1){list1.push_back(node[l1]);l1=node[l1].addressPost;}while(l2!=-1){list2.push_back(node[l2]);l2=node[l2].addressPost;}for(i=0,k=0;i<list1.size();i++){for(j=0;j<list2.size();j++){if(list1[i].addressPre==list2[j].addressPre){k=1;break;}}if(k)break;}if(k)cout <<setfill('0')<< setw(5)<<list1[i].addressPre<<endl;elsecout<<-1<<endl;}return 0; }

转载于:https://www.cnblogs.com/clnchanpin/p/7202242.html

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