1032 Sharing (25分) 我的代码和别人的代码-白红宇

1032 Sharing (25分) 我的代码和别人的代码

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发布时间：2019-05-23

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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).

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

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.

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:

Sample Input 2:

00001 00002 400001 a 1000110001 s -100002 a 1000210002 t -1

Sample Output 2:

-1

我用的是map

最后一个点超时，

22分代码：

#include
   
    #include
    #include
     
      #include
      
       using namespace std;struct node{	string add;	char d;	string next;};int main(){	map
       
         a;	vector
        
          a1,b1; int i,j,t,k,n,l1,l2; node temp; string s1,s2; map
         
          ::iterator it; a.clear(); a1.clear(); b1.clear(); cin>>s1>>s2>>n; for(i=0;i
          
           >temp.add>>temp.d>>temp.next; a.insert(make_pair(temp.add,temp)); } while(s1!="-1") { it=a.find(s1); a1.push_back(it->second); s1=it->second.next; } while(s2!="-1") { it=a.find(s2); b1.push_back(it->second); s2=it->second.next; }// for(i=0;i

但是还用更简单的：

#include 
   
    using namespace std;struct {    char key;    int next;    bool flag;}node[100010];int main(){    int d1,d2,n;    cin>>d1>>d2>>n;    for(int i=0;i
    
     >t1>>ch>>t2;        node[t1]={ch,t2,false};    }    for(int i=d1;i!=-1;i=node[i].next) node[i].flag=true;    for(int i=d2;i!=-1;i=node[i].next) {        if(node[i].flag){            printf("%05d",i);            return 0;        }    }    printf("-1");    return 0;}

这里需要注意：

node[t1]={ch,t2,false};

和你定义的顺序对应上。

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