比较考察技术含量的一道题。
参考链接:http://blog.csdn.net/lyy289065406/article/details/6647445
题目链接:http://poj.org/problem?id=2513
首先差不多能想到这事欧拉路，然后发现没法构图。没有尝试使用map，刚好最近在学字典树就直接上了。
然后就是并查集判断图连通。
参考链接里面的：连通图判断（并查集）+字典树（映射字符串）+欧拉路（奇数节点0个或2个）
注意：最后需要销毁字典树所创建的节点，不然会RE！
代码：先字典树，然后并查集，然后直接主函数中判断欧拉路
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cmath> 5 #include <cstring> 6 #include <string> 7 using namespace std; 8 #define maxm 500055 9 #define maxn 500055 10 #define maxnode 30 11 12 struct trie { 13 trie * next[maxnode]; 14 int id; 15 }; 16 trie *root; 17 int totid = 1; 18 trie * newnode() { 19 trie * t; 20 t = (trie*)malloc(sizeof(trie)); 21 t->id = 0; 22 for (int i = 0; i < maxnode; i++) 23 t->next[i] = 0; 24 return t; 25 } 26 void init() { 27 root = (trie *)malloc(sizeof(trie)); 28 root->id = 0; 29 for (int i = 0; i < maxnode; i++) 30 root->next[i] = 0; 31 } 32 int insert_find_Str(char s[]) { 33 trie *p = root, *q; 34 int len = strlen(s), i = 0; 35 while (i < len && p->next[s[i] - 'a']) { 36 p = p->next[s[i] - 'a']; 37 i++; 38 } 39 if (i == len && p->id) return p->id; 40 else if (i == len) return p->id = totid++; 41 while (i < len) { 42 q = newnode(); 43 p->next[s[i] - 'a'] = q; 44 p = p->next[s[i] - 'a']; 45 i++; 46 } 47 p->id = totid++; 48 return p->id; 49 } 50 void deltrie(trie* node) { 51 while (node) { 52 for (int i = 0; i < 26; i++) { 53 if ((node->next)[i]) 54 deltrie((node->next)[i]); 55 } 56 free(node); 57 node = 0; 58 } 59 } 60 61 int f[maxn]; 62 int deg[maxn]; 63 char st[maxn], anost[maxn]; 64 65 void init_ds(int n) { 66 for (int i = 1; i < n - 1; i++) 67 f[i] = i; 68 } 69 int find(int x) { 70 if (x == f[x]) 71 return f[x]; 72 return f[x] = find(f[x]); 73 } 74 void unite(int x, int y) { 75 int fx = find(x); 76 int fy = find(y); 77 if (fx == fy) return; 78 f[fx] = fy; 79 } 80 81 int main() { 82 init(); 83 init_ds(maxn); 84 memset(deg, 0, sizeof deg); 85 while (scanf("%s", st) > 0) { 86 scanf("%s", anost); 87 int a = insert_find_Str(st); 88 int b = insert_find_Str(anost); 89 deg[a]++; deg[b]++; 90 unite(a, b); 91 } 92 if (totid == 1) { 93 printf("Possible "); 94 return 0; 95 } 96 bool flag = true; 97 for (int i = 2; i < totid; i++) { 98 if (find(i) != find(i - 1)) { 99 flag = false; 100 break; 101 } 102 } 103 if (!flag) { 104 printf("Impossible "); 105 } 106 else { 107 int oddnum = 0; 108 for (int i = 1; i < totid; i++) 109 if (deg[i] & 1) oddnum++; 110 if (oddnum == 0 || oddnum == 2) { 111 printf("Possible "); 112 } 113 else 114 printf("Impossible "); 115 } 116 deltrie(root); 117 }
题目：
Colored Sticks
| Time Limit: 5000MS | Memory Limit: 128000K | |
| Total Submissions: 37263 | Accepted: 9775 |
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.