欧拉回路入门学习

POJ 1300 ZOJ 1395 Door Man 欧拉回路的判定

输入比较麻烦，其它感觉就是好水哦

#include<stdio.h>
#include<string.h>
int main()
{
    int i,j;
    char buf[128];
    int n,m;
    int door[22];
    while(gets(buf)!=NULL)
    {
        if(!strcmp(buf,"ENDOFINPUT"))break;
        if(buf[0]=='S')
        {
            sscanf(buf,"%*s %d %d",&m,&n);
            memset(door,0,sizeof(door));
            int doors=0;
            for(i=0;i<n;i++)
            {
                gets(buf);
                int k=0;
                while(sscanf(buf+k,"%d",&j)==1)
                {
                    doors++;
                    door[i]++;
                    door[j]++;
                    while(buf[k]&&buf[k]!=' ')k++;
                    while(buf[k]&&buf[k]==' ')k++;
                }
            }
            gets(buf);
            int odd=0,even=0;
            for(i=0;i<n;i++)
            {
                if(door[i]%2==1)odd++;
                else even++;
            }
            if(odd==0 && m == 0 || odd==2 && (door[m]%2) && (door[0]%2) && m)
                printf("YES %d\n",doors);
            else puts("NO");
        }
    }
}

 POJ 1386 ZOJ 2016 Play on Words

先判断是否满足欧拉回路的条件，用到并查集来判断是否连通（即是否只有一课树，题目要求一棵树）

#include<stdio.h>
#include<string.h>
#define INF 100000000
#define maxn 100001
int od[26],id[26];//出度，入度
int vis[26];
int f[26];
int n;
char word[1001];
struct node
{
    int u, v;
}edge[maxn];
int find(int x)
{
    int rt;
    for(rt = x;f[rt] >= 0;rt = f[rt]);
    while(rt != x)
    {
        int tmp = f[x];
        f[x] = rt;
        x = tmp;
    }
    return rt;
}
void merge(int x, int y)
{
     int rx = find(x);
     int ry = find(y);
     int tmp = f[rx] + f[ry];
     if(f[rx] < f[ry])
     {
         f[ry] = rx;
         f[rx] = tmp;
     }
     else
     {
         f[rx] = ry;
         f[ry] = tmp;
     }
}
bool connect()
{
    int i;
    memset(f,-1,sizeof(f));
    for(i=0;i<n;i++)
    {
        int u = edge[i].u ;
        int v = edge[i].v;
        if(u != v && find(u) != find(v))merge(u, v);
    }
    int tmp = -1;
    for(i=0;i<26;i++)
    {
        if(vis[i])
        {
            if(tmp == -1)tmp = i;
            else if(find(i) != find(tmp) )break;
        }
    }
    if(i==26)return true;
    return false;
}
int main()
{
    int u, v, i, j, cas;
    scanf("%d",&cas);
    while(cas--)
    {
        memset(od,0,sizeof(od));
        memset(id,0,sizeof(id));
        memset(vis,0,sizeof(vis));
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%s",word);
            int len=strlen(word);
            u = word[0]-'a'; v = word[len - 1] - 'a';
            od[u]++;id[v]++;
            vis[u]=vis[v]=1;
            edge[i].u = u;edge[i].v = v;
        }        
        bool ok = 1;
        int x = 0;
        int y = 0;
        for(i=0;i<26;i++)
        {
            if(!vis[i])continue;
            if(id[i] - od[i] >= 2 || od[i] - id[i] >= 2){ok=0;break;}
            if(od[i]==0 && id[i] == 0){ok=0;break;}
            if(id[i] - od[i] == 1)
            {
                x++;
                if(x>=2){ok=0;break;}
            }
            if(od[i] - id[i] == 1)
            {
                y++;
                if(y>=2){ok=0;break;}
            }
        }
        if(x != y)ok=0;
        if(!connect())ok=0;
        if(ok)puts("Ordering is possible.");
        else puts("The door cannot be opened.");
    }
    return 0;
}

 下面一个是完全自己写的 

#include<stdio.h>
#include<string.h>
#define maxn 100002
int fa[26];
bool vis[26];
int id[26],od[26];
int find(int x)
{
    int rt;
    for(rt = x ; fa[rt] != rt; rt = fa[rt]);
    while(x != rt)
    {
        int tmp = fa[x];
        fa[x] = rt;
        x = tmp;
    }
    return rt;
}
void merge(int x, int y)
{
    int rx = find(x);
    int ry = find(y);
    if(rx != ry)fa[rx] = ry;
}
int main()
{
    int cas, u, v,n;
    int i, j;
    char s[1002];
    scanf("%d",&cas);
    while(cas--)
    {
        scanf("%d",&n);
        memset(id,0,sizeof(id));
        memset(od,0,sizeof(od));
        memset(vis,0,sizeof(vis));
        for(i=0;i<26;i++)
            fa[i] = i;
        while(n--)
        {
            scanf("%s",s);
            int len =strlen(s);
            u = s[0]-'a'; v = s[len-1]-'a';
            od[u]++;id[v]++;
            vis[u] = vis[v] =1;
            if(u!=v)merge(u, v);
        }
        int tmp = -1;
        bool ok = true;
        for(i=0;i<26;i++)
        {
            if(vis[i])
            {
                if(tmp == -1)tmp = i;
                else if(find(i) != find(tmp)){ok=false;break;}
            }
        }
        if(!ok){printf("The door cannot be opened.\n");continue;}
        int x=0,y=0;
        for(i=0;i<26;i++)
        {
            if(vis[i])
            {
                if(id[i]==0 && od[i] == 0){ok=false;break;}
                if(id[i]-od[i]>=2||od[i]-id[i]>=2){ok=false;break;}
                if(id[i]-od[i]==1)
                {
                    x++;
                    if(x > 1){ok=false;break;}
                }
                if(od[i]-id[i]==1)
                {
                    y++;
                    if(y > 1){ok=false;break;}
                }
            }
        }       //printf("%d %d\n",id['m'-'a'],od['m'-'a']);
        if(ok)printf("Ordering is possible.\n");
        else printf("The door cannot be opened.\n");
    }
    return 0;
}