P1341 unordered letter pairs, P1341 letters

P1341 unordered letter pairs, P1341 letters
Description

Given n unordered letter pairs (case-sensitive, unordered, that is, the two letters in the letter pair can be reversed ). Create a string with n + 1 letters so that each letter pair appears in the string.

Input/Output Format Input Format:

Enter a positive integer n in the first line.

The following n rows contain two letters in each row, indicating that the two letters must be adjacent.

Output Format:

Output a string that meets the requirements.

If No string meets the requirements, output "No Solution ".

If there are multiple solutions, please output the solution with the smallest possible ASCII code (minimum Lexicographic Order) of the first letter

Input and Output sample Input example #1:

4aZtZXtaX

Output sample #1:

XaZtX 

Description

[Data scale and Conventions]

Different unordered letters have a limited number of pairs. The n size can be calculated.

 

 

This is the bare question of Euler's loop.

But there are two points to note:

1. output in a function cannot use a value-passing variable as a cyclic variable.

2. ios: sync may cause RE! ,

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<cstdlib> 6 using namespace std; 7 const int MAXN=4001; 8 void read(int & n) 9 {10     char c='+';int x=0;int flag=0;11     while(c<'0'||c>'9')12     {    c=getchar();    if(c=='-')    flag=1;    }13     while(c>='0'&&c<='9')14     {x=x*10+(c-48);c=getchar();}15     flag==1?n=-x:n=x;16 }17 int n; 18 int map[MAXN][MAXN];19 int indegree[MAXN];20 int ans[MAXN];21 int flag=0;22 void dfs(int num,int now)23 {24     ans[now]=num;25     if(now==n+1)26     {27         for(int i=1;i<=n+1;i++)28         printf("%c",(char)ans[i]);29         exit(0);30     }31     for(int i=65;i<=127;i++)32     {33         if(map[num][i])34         {35             map[num][i]=0;36             map[i][num]=0;37             dfs(i,now+1);38             map[num][i]=1;39             map[i][num]=1;40         }41     }42     ans[now]=0;43 }44 int main()45 {46     cin>>n;47     ios::sync_with_stdio(false);48     for(int i=1;i<=n;i++)49     {50         char a,b;51         cin>>a>>b;52         if(!map[a][b])53         {54             map[a][b]=1;55             map[b][a]=1;56             indegree[a]++;57             indegree[b]++;58         }59     }60     int tot=0;61     int bgji=0x7fff,bgou=0x7ffff;62     for(int i=65;i<=127;i++)63     {64         if(indegree[i]%2==1)65         {66             tot++;67             bgji=min(bgji,i);68         }69         else if(indegree[i])70         bgou=min(i,bgou);71     }72     if(tot!=0&&tot!=2)73     {74         printf("No Solution");75         exit(0);76     }77     tot==0?78     dfs(bgou,1):79     dfs(bgji,1);80     return 0;81 }