(字典树+欧拉通路) poj 2513 Colored Sticks

Colored Sticks

Time Limit: 5000MS		Memory Limit: 128000K
Total Submissions: 16750		Accepted: 4248

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 Input

blue redred violetcyan blueblue magentamagenta cyan

Sample Output

Possible

Hint

Huge input,scanf is recommended.

/* 题目大意：每个木棒头尾都有两种颜色,木棒末端颜色相同的可以连接成一条直线,问能否将所有木棒连接起来组成一条直线。 题目求解：传换成图的问题就是求解能否一笔画成（欧拉通路） 代码技巧：用数组字典树和链表字典树都可以，我看很多用链表的，那我就来一个静态的吧。 ps：没注意检查那个下标，真的郁闷死我了。调试半天 Source Code Problem: 2513 User: wawadimu Memory: 64224K Time: 938MS Language: C++ Result: Accepted Source Code */ #include<iostream> using namespace std; #define maxn 50001 int father[maxn];//i的父亲数组 int rank[maxn];//i所在集合的秩(树的高度) int degree[maxn];//节点的度数 struct node//字典树 { int key; int next[26]; node() { key=0;//这里0和-1都可以只要是没有用过的就可以. memset(next,-1,sizeof(next)); } }trie[600000];//其实不用这么多，但是也没超内存 unsigned int loc=2;//记录当前新节点 unsigned int num=1;//单词数目 int insert(char *str)//字典树的插入操作 { int index,r=1; int len=strlen(str); for(int i=0;i<len;i++) { index=str[i]-'a';//最初这个用'A',没有注意到颜色使用小写字母的,本地运算由于数据居然没有报错，调试很半天才发现.晕 if(trie[r].next[index]==-1) { trie[r].next[index]=loc++;//尾插法,nex[index]记录index的下一个节点坐标 r=loc-1; } else { r=trie[r].next[index]; } } if(trie[r].key==0)//字典树单词不存在 { trie[r].key=num++; } return trie[r].key; } void _make_set(int x)//开始每一个i都是一个独立的集合 { father[x]=x; degree[x]=0; rank[x]=0; } int _find_set(int x)//查找x的所在集合，并递归压缩路径 { if(x!=father[x]) { father[x]=_find_set(father[x]); } return father[x]; } void _union(int x,int y)//合并两个集合, { if(x==y) return; if(rank[x]>rank[y])//将秩少的合并到秩多的根结点,避免退化成线性 { father[y]=x; } else { if(rank[y]==rank[x]) rank[y]++; father[x]=y; } } int main() { //freopen("2513.txt","r",stdin); int i,j; char s1[11], s2[11]; for (i = 0; i < maxn; i++) { _make_set(i); } while(scanf("%s%s",s1,s2)!=EOF) { //cout<<s1<<" "<<s2<<endl; int x=insert(s1); int y=insert(s2); degree[x]++; degree[y]++; //cout<<x<<" "<<y<<endl; _union(_find_set(x),_find_set(y)); //cout<<father[x]<<endl; } int sum=0; num--; //cout<<num<<endl; for(i=1;i<=num;i++) if(degree[i]%2) sum++;// // cout<<sum<<endl; //存在欧拉通路的条件:奇顶点数为0或者2，并且图是连通图. if(sum>2) printf("Impossible/n");//奇顶点 else { for(j=2;j<=num;j++)//检查是否为连通图 { if(_find_set(j)!=_find_set(1)) { printf("Impossible/n"); return 1; } } printf("Possible/n"); } return 1; }