POJ 2513

题目链接：

http://poj.org/problem?id=2513

http://bailian.openjudge.cn/practice/2513?lang=en_US

Time Limit: 5000MS Memory Limit: 128000K

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.

题意：

给出若干木棒，每个木棒两个端点有两个颜色，两个木棒的端点颜色相同，则可以接在一起。问是否把所有木棒接成一根。

题解：

（参考http://blog.csdn.net/lyy289065406/article/details/6647445）

将每个颜色都看成一个节点，木棒就能看成一条连接两个端点的无向边。问题就变成在给定一个无向图问是否存在欧拉路。

欧拉路存在的充要条件是：① 图是连通的； ② 所有节点的度为偶数，或者有仅有两个度数为奇数的节点。

这道题，求顶点的度数很好求，但是怎么判断图是否连通呢？通常来说，DFS/BFS或者并查集都能做。

不过，本题字典树的用处就是代替map，将输入字符串快速hash成一个值。

AC代码（BFS判图的连通性）：

#include<bits/stdc++.h>
using namespace std;
const int maxn=250000+10;

int degree[2*maxn];
vector<int> G[2*maxn];

namespace Trie
{
    const int SIZE=maxn*10;
    int sz;
    int idtot;
    struct TrieNode{
        int ed;
        int nxt[26];
    }trie[SIZE];
    void init()
    {
        idtot=0;
        sz=1;
    }
    int insert(const string& s)
    {
        int p=1;
        for(int i=0;i<s.size();i++)
        {
            int ch=s[i]-'a';
            if(!trie[p].nxt[ch]) trie[p].nxt[ch]=++sz;
            p=trie[p].nxt[ch];
        }
        if(!trie[p].ed) trie[p].ed=++idtot;
        return trie[p].ed;
    }
};

bool vis[2*maxn];
int bfs(int s)
{
    int res=0;
    memset(vis,0,sizeof(vis));
    queue<int> Q;
    Q.push(s), vis[s]=1, res++;
    while(!Q.empty())
    {
        int u=Q.front(); Q.pop();
        for(int i=0;i<G[u].size();i++)
        {
            int v=G[u][i];
            if(!vis[v]) Q.push(v), vis[v]=1, res++;
        }
    }
    return res;
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    string s1,s2;
    Trie::init();
    while(cin>>s1>>s2)
    {
        int u=Trie::insert(s1), v=Trie::insert(s2);
        G[u].push_back(v);
        G[v].push_back(u);
        degree[u]++;
        degree[v]++;
    }

    int cnt=0;
    for(int i=1;i<=Trie::idtot;i++) {
        if(degree[i]%2) cnt++;
    }
    if(cnt==1 || cnt>2) cout<<"Impossible
";
    else
    {
        if(bfs(1)<Trie::idtot) cout<<"Impossible
";
        else cout<<"Possible
";
    }
}

AC代码（并查集判图的连通性）：

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=250000+5;

namespace Trie
{
    const int SIZE=maxn*10;
    int sz;
    int idcnt;
    struct TrieNode{
        int ed;
        int nxt[26];
    }trie[SIZE];
    void init()
    {
        idcnt=0;
        sz=1;
    }
    int insert(char *s)
    {
        int len=strlen(s), p=1;
        for(int i=0;i<len;i++)
        {
            int ch=s[i]-'a';
            if(!trie[p].nxt[ch]) trie[p].nxt[ch]=++sz;
            p=trie[p].nxt[ch];
        }
        return (trie[p].ed)?(trie[p].ed):(trie[p].ed=++idcnt);
    }
};

int par[2*maxn],ran[2*maxn];
int find(int x){return (par[x]==x)?x:(par[x]=find(par[x]));}
void unite(int x,int y)
{
    x=find(x), y=find(y);
    if(x==y) return;
    if(ran[x]<ran[y]) par[x]=y;
    else par[y]=x, ran[x]+=(ran[x]==ran[y]);
}

int degree[2*maxn];
int main()
{
    Trie::init();
    memset(degree,0,sizeof(degree));
    for(int i=1;i<2*maxn;i++) par[i]=i,ran[i]=0;

    char s[2][13];
    while(scanf("%s %s",s[0],s[1])!=EOF)
    {
        int u=Trie::insert(s[0]);
        int v=Trie::insert(s[1]);
        if(find(u)!=find(v)) unite(u,v);
        degree[u]++;
        degree[v]++;
    }

    int cnt=0;
    int z=find(1);
    for(int i=1;i<=Trie::idcnt;i++)
    {
        if(find(i)!=z) {
            printf("Impossible
");
            return 0;
        }

        if(degree[i]%2) cnt++;
        if(cnt>2) {
            printf("Impossible
");
            return 0;
        }
    }
    if(cnt==1) {
        printf("Impossible
");
        return 0;
    } else {
        printf("Possible
");
        return 0;
    }
}