nyoj 230/poj 2513 彩色棒 并查集+字典树+欧拉回路

题目链接：http://acm.nyist.net/JudgeOnline/problem.php?pid=230

题意：给你许许多多的木棍，没条木棍两端有两种颜色，问你在将木棍相连时，接触的端点颜色必须相同，是否能把它们都连起来

思路：很明显的欧拉路径，但题目给的字符串数据很大，得用字典树存取。

代码如下：

#include "stdio.h"
#include "string.h"
#include "stdlib.h"

#define N 505000
int set[N],du[N];

int find(int x)
{
    if(set[x]==-1) return x;
    return set[x]=find(set[x]);
}
void bing(int a,int b)
{
    int fa = find(a);
    int fb = find(b);
    if(fa!=fb) set[fa] = fb;
}

struct node
{
    struct node *next[26];
    int num;
};
int id;

void BFS(node *root);
int Trie(node *root,char *s);
void Date_process(int n)
{
    int i=0;
    int t1,t2;
    char s1[15],s2[15];
    node *root = (node *)malloc(sizeof(node));
    for(i=0; i<26; ++i)
        root->next[i] = NULL;
    root->num = -1;
    id = 0;
    memset(du,0,sizeof(du));
    memset(set,-1,sizeof(set));

    for(i=0; i<n; ++i)
    {
        scanf("%s %s",s1,s2);
        t1 = Trie(root,s1);
        t2 = Trie(root,s2);
        du[t1]++;
        du[t2]++;
        bing(t1,t2);
    }
    BFS(root);
}

int main()
{
    int T;
    int i,n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        getchar();
        if(n==0)
        {
            printf("Possible
");
            continue;
        }
        Date_process(n);
        bool flag = true;
        int k = find(0);
        for(i=0; i<id; ++i)
        {
            if(find(i)!=k)
                flag = false;
        }
        if(!flag)  //若该图不连通，Impossible
        {
            printf("Impossible
");
            continue;
        }
        int ans=0;
        for(i=0; i<id; ++i)
        {
            if(du[i]%2==1)
                ans++;
        }
        if(ans==2 || ans==0)  //欧拉回路或者欧拉路径 Possible
            printf("Possible
");
        else
            printf("Impossible
");
    }
    return 0;
}

int Trie(node *root,char *s)  //字典树
{
    int i=0;
    node *p=root;
    while(s[0]!='')
    {
        if(p->next[s[0]-'a']==NULL)
        {
            node *tt;
            tt = (node *)malloc(sizeof(node));
            tt->num = -1;
            for(i=0; i<26; ++i) tt->next[i] = NULL;
            p->next[s[0]-'a'] = tt;
            p = tt;
        }
        else
            p = p->next[s[0]-'a'];
        s++;
    }
    if(p->num==-1)
        p->num = id++;
    return p->num;
}

void BFS(node *root)  //深搜释放内存
{
    int i;
    if(root==NULL) return ;
    for(i=0; i<26; i++)
    {
        if(root->next[i]!=NULL)
            BFS(root->next[i]);
    }
    free(root);
}