POJ 2513 Colored Sticks

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这题是求欧拉路径，给出字符串，map映射会超时，用Trie树来映射。
一个图有欧拉路径的充分必要条件是这个图是连通的且每一个顶点的度都是偶数或者且仅有两个顶点的度是奇数。联通可以用并查集来判断。

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;int fa[510000],trie[510000][30],dev[510000],id[510000],sz,num;int getfa(int x){    return fa[x] == x ? x : fa[x] = getfa(fa[x]);}void init(){    for(int i = 0; i < 510000; i++)        fa[i] = i;    memset(trie,0,sizeof(trie));    memset(dev,0,sizeof(dev));    memset(id,0,sizeof(id));    sz = 1;    num = 0;}int Insert(char *s){    int cur = 1;    for(int i = 0; s[i]; i++)    {        int c = s[i] - 'a';        if(!trie[cur][c])        {            trie[cur][c] = ++sz;        }        cur = trie[cur][c];    }    dev[cur]++;    if(dev[cur] == 1) id[cur] = ++num;    return id[cur];}int main(){    char s1[15],s2[15];    init();    while(scanf("%s%s",s1,s2)!=EOF)    {        int a = Insert(s1);        int b = Insert(s2);        //printf("%s %s\n",s1,s2);        int x = getfa(a);        int y = getfa(b);        if(x != y) fa[y] = x;    }    if(num == 0)    {        printf("Possible\n");        return 0;    }    int count1 = 0;    for(int i = 1; i <= sz; i++)    {        if(dev[i]&1) count1 ++;//计算度为奇的点的个数。        if(count1 > 2) break;    }    int count2 = 0;    for(int i = 1; i <= num; i++)        if(fa[i] == i) count2++;//count2 == 1说明联通。    if(count2 == 1 && (count1 == 0 || count1 == 2)) printf("Possible\n");    else printf("Impossible\n");    return 0;}

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