POJ 2513 Colored Sticks (Trie tree + parallel query set + Euler's path), pojtrie

POJ 2513 Colored Sticks (Trie tree + parallel query set + Euler's path), pojtrie

Colored Sticks

Time Limit:5000 MS		Memory Limit:128000 K
Total Submissions:31490		Accepted:8320

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 redred violetcyan blueblue magentamagenta cyan

Sample Output

Possible

Hint

Huge input, scanf is recommended.

Source

The UofA Local 2000.10.14

Question link: http://poj.org/problem? Id = 2513

Give some wooden sticks with colors on both ends. Only the two ends of the sticks have the same color.

Question analysis: the question is not difficult. A typical undirected graph determines whether there is an Euler's path or loop problem, but there are too many strings and map times out. Here we use the Trie tree to give each wooden stick label, you can determine whether there is an Euler's path or loop to catch two points. The first is graph connectivity, and the second is that there can only be zero (loop) or two (PATH) odd-degree nodes)

#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");}