POJ 2513 Colored Sticks (Trie树＋并查集＋欧拉路)

Colored Sticks

Description

You are given a bunch of wooden sticks. Each endpoint of each stick is colored with some color. Is it possible to align the sticks in a straight line such that the colors of the endpoints that touch
are of the same color?

Input

Input is a sequence of lines, each line contains two words, separated by spaces, giving the colors of the endpoints of one stick. A word is a sequence of lowercase letters no longer than 10 characters.
There is no more than 250000 sticks.

Output

If the sticks can be aligned in the desired way, output a single line saying Possible, otherwise output Impossible.

Sample Input

blue red
red violet
cyan blue
blue magenta
magenta cyan

Sample Output

Possible

Hint

Huge input,scanf is recommended.

Source

The UofA Local 2000.10.14

题目链接：http://poj.org/problem?id=2513

题目大意：给一些木棍，两端都有颜色，只有两根对应的端点颜色相同才能相接，问能不能把它们接成一根木棍

题目分析：题意不难，典型的无向图判断是否存在欧拉通路或回路的问题，但是字符串太多，用map超时，这里使用Trie树给每个木棍标号，判断是否存在欧拉通路或回路抓住两点就行了，第一是图连通，第二是奇数度结点只能有0个(回路)或2个(通路)

#include <cstdio>
#include <cstring>
int const MAX = 500005;
int fa[MAX], d[MAX], cnt;

struct Trie
{
    int sz, t[MAX][15];
    int jud[MAX];
    Trie()
    {
        sz = 1;
        memset(t[0], -1, sizeof(t));
        jud[0] = 0;
    }
    void clear()
    {
        sz = 1;
        memset(t[0], -1, sizeof(t));
        jud[0] = 0;
    }
    int idx(char c)
    {
        return c - 'a';
    }
    void insert(char* s, int v)
    {
        int u = 0, len = strlen(s);
        for(int i = 0; i < len; i++)
        {
            int c = idx(s[i]);
            if(t[u][c] == -1)
            {
                memset(t[sz], -1, sizeof(t[sz]));
                jud[sz] = 0;
                t[u][c] = sz++;
            }
            u = t[u][c];
        }
        jud[u] = v;
    }
    int search(char* s)
    {
        int u = 0, len = strlen(s);
        for(int i = 0; i < len; i++)
        {
            int c = idx(s[i]);
            if(t[u][c] == -1)
                return -1;
            u = t[u][c];
        }
        if(jud[u])
            return jud[u];
        return -1;
    }
}t;

void Init()
{
    for(int i = 0; i < MAX; i++)
        fa[i] = i;
}

int Find(int x)
{
    return x == fa[x] ? x : fa[x] = Find(fa[x]);
}

void Union(int a, int b)
{
    int r1 = Find(a);
    int r2 = Find(b);
    if(r1 != r2)
        fa[r1] = r2;
}

bool eluer()
{
    int sum = 0, t = -1;
    for(int i = 1; i < cnt; i++)
        if(d[i] % 2)
            sum++;
    if(sum != 0 && sum != 2)
        return false;
    for(int i = 1; i < cnt; i++)
    {
        if(t == -1)
            t = Find(i);
        else if(Find(i) != Find(t))
            return false;
    }
    return true;
}

int main()
{
    char s1[20],s2[20];
    cnt = 1;
    Init();
    t.clear();
    while(scanf("%s %s", s1, s2) != EOF)
    {
        if(t.search(s1) == -1)
            t.insert(s1, cnt++);
        int u = t.search(s1);
        if(t.search(s2) == -1)
            t.insert(s2, cnt++);
        int v = t.search(s2);
        Union(u, v);
        d[u]++;
        d[v]++;
    }
    if(eluer())
        printf("Possible\n");
    else
        printf("Impossible\n");
}