Colored Sticks
| Time Limit: 5000MS | Memory Limit: 128000K | |
| Total Submissions: 24844 | Accepted: 6542 |
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
1 #include<stdio.h> 2 #include<string.h> 3 #include<stdlib.h> 4 typedef struct node 5 { 6 int order; //存储该颜色在对应数组中的编号 7 struct node *next[26]; 8 }node; 9 10 //注意最多有250000个棒，所以最多应该有500000种颜色 11 int father[500005]; //并查集中记录父亲节点 12 int degree[500005]; //记录每个颜色出现的次数，用于欧拉路的判断 13 14 int t,k; //t为记录标号增加的数组指针，k为静态内存分配时的数组指针 15 16 int search(char *s,node *T) //返回s所表示的颜色在数组中的编号 17 { 18 int len,i,j,id,flag=0; 19 node *p,*q; 20 len=strlen(s); 21 p=T; 22 for(i=0;i<len;++i) 23 { 24 id=s[i]-'a'; 25 if(p->next[id]==NULL) 26 { 27 flag=1; 28 q=(node *)malloc(sizeof(node)); 29 for(j=0;j<26;++j) 30 q->next[j]=NULL; 31 p->next[id]=q; 32 } 33 p=p->next[id]; 34 } 35 if(flag) //肯定是新出现的颜色---因为有新开辟的节点 36 { 37 p->order=t; //t存储该颜色在数组中对应的编号 38 degree[t++]++; //该颜色的度增加一 39 return p->order; 40 } 41 else 42 { 43 if(p->order==0) //也是新颜色----因为没有颜色在此标记 44 { 45 p->order=t; //t存储该颜色在数组中对应的编号 46 degree[t++]++; //该颜色的度增加一 47 return p->order; 48 } 49 degree[p->order]++; //出现过的颜色，对应的度增加一 50 return p->order; 51 } 52 } 53 54 int find_father(int i) 55 { 56 if(father[i]==-1) 57 return i; 58 else 59 return find_father(father[i]); 60 } 61 62 void merge(int num1,int num2) 63 { 64 num1=find_father(num1); 65 num2=find_father(num2); 66 if(num1!=num2) 67 { 68 if(father[num1]<father[num2]) //father[num1]的绝对值大 69 father[num2]=num1; 70 else 71 father[num1]=num2; 72 } 73 } 74 75 int main() 76 { 77 int i,num1,num2,flag; 78 char s1[20],s2[20]; 79 node *T; 80 t=1; //从1号开始算起 81 flag=0; 82 memset(father,-1,sizeof(father)); 83 memset(degree,0,sizeof(degree)); 84 T=(node *)malloc(sizeof(node)); 85 T->order=0; 86 for(i=0;i<26;++i) 87 T->next[i]=NULL; 88 while(scanf("%s%s",s1,s2)!=EOF) 89 { 90 num1=search(s1,T); 91 num2=search(s2,T); 92 merge(num1,num2); 93 } 94 for(i=1;i<t;++i) //共有t-1种颜色 95 if(father[i]==-1) 96 ++flag; 97 if(flag>1) //说明不连通 98 printf("Impossible\n"); 99 else 100 { 101 flag=0; 102 for(i=1;i<t;++i) 103 if(degree[i]&1) //统计奇数度结点出现的次数 104 ++flag; 105 if(flag==2||flag==0) //无向图G具有一条欧拉路，当且仅当G是连通的，且有零个或两个奇数度结点 106 printf("Possible\n"); 107 else 108 printf("Impossible\n"); 109 } 110 return 0; 111 }