洛谷1341 无续字母对

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
int g[55][55], deg[55];
vector<int > ans;
int tonum(char x){                                               //字符转为数字
    if(x<=‘Z‘) return x-‘A‘+1;
    return x-‘a‘+1+26;
}
char tochar(int x){                                                //数字转字符
    if(x<=26) return x+‘A‘-1;
    return x+‘a‘-1-26;
}
void dfs(int here){                                                //从单点开始的欧拉回/链路dfs
    for(int i=1;i<=52;i++)                                  //确保字典序
        if(g[here][i]){
            g[here][i]=0;                                 //删除边线
            g[i][here]=0;
            dfs(i);
        }
    ans.push_back(here);                                   //添加顶点到回/链路中
}
int main(){
    int n, u1=100, u2=100, cnt=0;
    cin>>n;
    for(int i=1;i<=n;i++){                                       //建图
        char x, y;
        cin>>x>>y;
        if(!g[tonum(x)][tonum(y)]){
            g[tonum(x)][tonum(y)]=1;
            g[tonum(y)][tonum(x)]=1;
            deg[tonum(x)]++;                              //出入度统计
            deg[tonum(y)]++;
        }
    }
    for(int i=1;i<=52;i++)                                                                     //统计奇点数
        if(deg[i]&1){                          
            cnt++;
            u1=min(u1, i);                                                                     //满足字典序
        }
        else if(deg[i]!=0) u2=min(u2, i);
    if(cnt!=0&&cnt!=2){ cout<<"No Solution"<<endl; return 0; }             //奇点数不满足（2）
    if(cnt==0) dfs(u2);
    else dfs(u1);
    reverse(ans.begin(), ans.end());                                                             //反转回/链路
    if(ans.size()>n+1){ cout<<"No Solution"<<endl; return 0; }                //代表不连通图情况（1）
    for(int i=0;i<ans.size();i++)
        cout<<tochar(ans[i]);
    cout<<endl;
    return 0;
}