第六章部分例题

#include <iostream>
#include <cstdio>
#include <set>
#include <cstring>

using namespace std;

const int maxn=1000;
int UG[maxn][maxn];
int vis[maxn];
char str[1024];
int in[maxn];
int out[maxn];
set<int> s;
unsigned dfs_num=0;

void dfs(int u)
{
    vis[u]=1;
    dfs_num++;

    for(int i=0;i<='z'-'a';i++)
        if(UG[u][i])
            if(!vis[i])
                dfs(i);
}

int main()
{
    s.clear();
    memset(UG,0,sizeof(UG));
    memset(vis,0,sizeof(vis));
    memset(in,0,sizeof(in));
    memset(out,0,sizeof(out));

    int n;

    scanf("%d",&n);

    while(n--)
    {
        memset(str,0,sizeof(str));
        scanf("%s",str);

        printf("----%s---
",str);


        if(s.count(str[0]-'a')==0)
            s.insert(str[0]-'a');
        if(s.count(str[strlen(str)-1]-'a')==0)
            s.insert(str[strlen(str)-1]-'a');

        UG[str[0]-'a'][str[strlen(str)-1]-'a']++;
        UG[str[strlen(str)-1]-'a'][str[0]-'a']++;

        out[str[0]-'a']++;
        in[str[strlen(str)-1]-'a']++;
    }

    //find odd
    
    //if have odd out return it
    //if have odd in return it
    
    printf("%d
",s.size());

    int num1=0;
    int num2=0;

    for(int i=0;i<='z'-'a';i++)
    {
        i==0? printf("in:%d",in[i]):printf(" %d",in[i]);
    }

    cout<<endl;

    for(int i=0;i<='z'-'a';i++)
        i==0? printf("out:%d",out[i]):printf(" %d",out[i]);

    cout<<endl;


    //计算多少个入度不等于出度
    for(int i=0;i<'z'-'a';i++)
        if(in[i]!=out[i]) num1++;


    //大于两个就有问题
    if(num1>2)
    {
        cout<<"NO"<<endl;
        return 0;
    }


    //有多少个入度不等于出度的0个?
    
    
    if(num1==0)
    {
        for(int i=0;i<'z'-'a';i++)
        {
            if(s.count(i))
            {
                dfs(i);

                if(s.size()==dfs_num)
                {
                    cout<<"YES"<<endl;

                    return 0;
                }
                else
                {
                    cout<<"NO"<<endl;

                    return 0;
                }
            }
        }
    }


    //大于两个,进一步判断是否满足条件
    if(num1==2)
    {
        for(int i=0;i<'z'-'a';i++)
            if(in[i]!=out[i])
            {
                num2=num1;
                num1=i;
            }

            printf("num1:%d
",num1);
            printf("num2:%d
",num2);

            if((in[num1]==out[num1]+1 && in[num2]+1==out[num2]) || (in[num2]==out[num2]+1 && in[num1]+1==out[num1]))
            {
                for(int i=0;i<'z'-'a';i++)
                    if(s.count(i))
                    {

                        //标记所有节点判断是否连通
                        dfs(i);

                        if(s.size()==dfs_num) cout<<"YES"<<endl;
                        else 
                            cout<<"NO"<<endl;

                        return 0;
                    }
            }
            else 
                cout<<"NO"<<endl;
    }

    return 0;
}