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 505000int 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\n");            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\n");            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\n");        else            printf("Impossible\n");    }    return 0;}int Trie(node *root,char *s)  //字典树{    int i=0;    node *p=root;    while(s[0]!='\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);}