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
题意:给你n(n<=250000)根小木棒,木棒两端染有一些颜色,两根木棒相连需要保证相连端点颜色一致.求这些木棒是否可以全部相连?
题解:可以将木棒理解为边,连接两种颜色,那么全部相连就变成了一笔画的问题,即判断欧拉回路.
欧拉回路需要满足两个条件:
1.度数为奇数的点为0个或2个
2.图联通
现在的问题是如何把字符串变成数字的点,用map是肯定会TLE的
hash应该可以,但tire好写,所以我选择了trie树,存入trie的值为该字符串的映射
图联通用并查集去判断.
代码如下:
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; int cnt=0,du[500010],fa[500050]; struct Trie { int sz,tr[500010][26]; int time[500010]; void clear() { sz=0; memset(tr,0,sizeof(tr)); } void insert(char* c,int x) { int rt=0,len=strlen(c); for(int i=0;i<len;i++) { if(!tr[rt][c[i]-‘a‘]) { tr[rt][c[i]-‘a‘]=++sz; } rt=tr[rt][c[i]-‘a‘]; } time[rt]=x; } int search_t(char *c) { int rt=0,len=strlen(c); for(int i=0;i<len;i++) { if(!tr[rt][c[i]-‘a‘]) { return 0; } rt=tr[rt][c[i]-‘a‘]; } return time[rt]; } }trie; void mem(int n) { for(int i=1;i<=n;i++) { fa[i]=i; } } int find(int x) { if(x!=fa[x]) { fa[x]=find(fa[x]); } return fa[x]; } void union_(int x,int y) { int fx=find(x); int fy=find(y); if(fx!=fy) { fa[fx]=fy; } } int euler() { int sum=0,t=find(1); for(int i=1;i<=cnt;i++) { if(du[i]%2==1) { sum++; } } if(sum!=0&&sum!=2) { return 0; } for(int i=1;i<=cnt;i++) { if(find(i)!=find(t)) { return 0; } } return 1; } int main() { char s1[20],s2[20]; mem(500000); // freopen("1.in","r",stdin); // freopen("1.out","w",stdout); while(scanf("%s %s\n",s1,s2)!=EOF) { int from,to; if(!trie.search_t(s1)) { trie.insert(s1,++cnt); } from=trie.search_t(s1); du[from]++; if(!trie.search_t(s2)) { trie.insert(s2,++cnt); } to=trie.search_t(s2); du[to]++; union_(from,to); } if(euler()) { puts("Possible"); } else { puts("Impossible"); } }
