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

Colored Sticks

Time Limit: 5000MS Memory Limit: 128000KTotal 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

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");}