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\n");
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\n");
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\n");
112 }
113 else
114 printf("Impossible\n");
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.