问题便转化为：给定一个图，是否存在“一笔画”经过涂中每一点，以及经过每一边一次。这样就是求图中是否存在欧拉路Euler-Path。
由图论知识可以知道，无向图存在欧拉路的充要条件为：
① 图是连通的；
② 所有节点的度为偶数，或者有且只有两个度为奇数的节点。
trie 和 并查集
1 #include <iostream> 2 #include <vector> 3 #include <string.h> 4 #include<stdlib.h> 5 #include<stdio.h> 6 using namespace std; 7 typedef struct trie{ 8 int trie_index; 9 bool flag; 10 trie* next[26]; 11 12 } Trie; 13 void trie_init(Trie* tr){ 14 tr->trie_index = -1; 15 tr->flag = false; 16 for(int i = 0; i< 26; i ++) 17 tr->next[i] = NULL; 18 } 19 int trie_index; 20 Trie *root; 21 vector<int> trie_set; 22 vector<int> trie_rank; 23 24 int trie_hash(char *input){ 25 int i = 0; 26 Trie* path = root; 27 while(input[i] != 0){ 28 if(path->next[input[i] - 'a'] == NULL){ 29 path->next[input[i] - 'a'] = (Trie*)malloc(sizeof(Trie)); 30 trie_init(path->next[input[i] - 'a']); 31 } 32 path = path->next[input[i] - 'a']; 33 i ++; 34 } 35 if(path->flag == true){ 36 return path->trie_index; 37 }else{ 38 path->flag = true; 39 path->trie_index = trie_index ++; 40 return path->trie_index; 41 } 42 43 } 44 int find_set(int i){ 45 if(trie_set[i] == i) 46 return i; 47 return find_set(trie_set[i]); 48 49 } 50 void union_set(int i, int j){ 51 int i_p = find_set(i); 52 int j_p = find_set(j); 53 if(i_p == j_p) return; 54 if(trie_rank[i_p] > trie_rank[j_p]) 55 trie_set[j_p] = i_p; 56 else{ 57 if(trie_rank[i_p] == trie_rank[j_p]) 58 trie_rank[j_p]++; 59 trie_set[i_p] = j_p; 60 } 61 62 63 } 64 int main(int argc, char* argv[]){ 65 trie_index = 0; 66 vector<int> trie_degree(500000,0); 67 char a[12], b[12]; 68 root = (Trie*)malloc(sizeof(Trie)); 69 trie_init(root); 70 for(int i = 0; i < 500000; i ++){ 71 trie_set.push_back(i); 72 trie_rank.push_back(0); 73 } 74 while (scanf("%s%s",a,b)!=EOF){ 75 int i = trie_hash(a); 76 int j = trie_hash(b); 77 trie_degree[i] += 1; 78 trie_degree[j] += 1; 79 80 union_set(i, j); 81 82 } 83 int pre = find_set(0); 84 bool result = true; 85 for(int i = 1; i < trie_index; i ++) 86 if(find_set(i) != pre) 87 result = false; 88 int odd_num = 0; 89 for(int i = 0; i < trie_index; i ++){ 90 if((trie_degree[i] & 1) == 1) 91 odd_num += 1; 92 } 93 if(odd_num== 1 || odd_num > 2) 94 result = false; 95 if(result == true) 96 cout << "Possible" << endl; 97 else 98 cout << "Impossible" << endl; 99 return 0; 100 }
样本输入
Sample Input
blue red
red violet
cyan blue
blue magenta
magenta cyan
Sample Output
Possible