POJ-1386 Play on Words

Play on Words

Time Limit: 1000MS Memory Limit: 10000K   

Description

Some of the secret doors contain a very interesting word puzzle. The team of archaeologists has to solve it to open that doors. Because there is no other way to open the doors, the puzzle is very important for us. 

There is a large number of magnetic plates on every door. Every plate has one word written on it. The plates must be arranged into a sequence in such a way that every word begins with the same letter as the previous word ends. For example, the word ``acm'' can be followed by the word ``motorola''. Your task is to write a computer program that will read the list of words and determine whether it is possible to arrange all of the plates in a sequence (according to the given rule) and consequently to open the door. 

Input

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing a single integer number Nthat indicates the number of plates (1 <= N <= 100000). Then exactly Nlines follow, each containing a single word. Each word contains at least two and at most 1000 lowercase characters, that means only letters 'a' through 'z' will appear in the word. The same word may appear several times in the list.

Output

Your program has to determine whether it is possible to arrange all the plates in a sequence such that the first letter of each word is equal to the last letter of the previous word. All the plates from the list must be used, each exactly once. The words mentioned several times must be used that number of times. 
If there exists such an ordering of plates, your program should print the sentence "Ordering is possible.". Otherwise, output the sentence "The door cannot be opened.". 

Sample Input

32acmibm3acmmalformmouse2okok

Sample Output

The door cannot be opened.Ordering is possible.The door cannot be opened.

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思路：是欧拉回路还是欧拉通路呢？every word begins with the same letter as the previous word ends.只要求后一个字母的开头和前一个字母的末尾相同，因此是欧拉通路。

判定有向图的欧拉通路：

基图连通、所有顶点的出度、入度相等，或者起点的出度 - 入度 = 1，终点的入度 - 出度 = 1，其余相等。

代码如下：

/*ID: j.sure.1PROG:LANG: C++*//****************************************/#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <stack>#include <queue>#include <vector>#include <map>#include <string>#include <climits>#include <iostream>#define INF 0x3f3f3f3fusing namespace std;/****************************************/const int M = 1e5+5;int m, tot, head[26], fa[26];struct Node {int u, v, next;}edge[M];bool vis[26];int od[26], id[26];void add(int u, int v){edge[tot].u = u; edge[tot].v = v;edge[tot].next = head[u];head[u] = tot++;}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), fy = Find(y);if(fx != fy) {fa[fy] = fx;}}bool connect(){for(int i = 0; i < 26; i++) fa[i] = i;for(int i = 0; i < m; i++) {//遍历所有边int u = edge[i].u, v = edge[i].v;Union(u, v);}int root = -1;for(int i = 0; i < 26; i++) if(vis[i]) {if(root == -1) root = Find(i);else {if(Find(i) != root) return false;}}return true;}bool Euler(){int one = 0, none = 0;for(int i = 0; i < 26; i++) if(vis[i]) {if(abs(id[i] - od[i]) >= 2) return false;if(od[i] - id[i] == 1) {one++;if(one > 1) return false;}if(id[i] - od[i] == 1) {none++;if(none > 1) return false;}}if(one != none) return false;return true;}int main(){#ifdef J_Sure//freopen("000.in", "r", stdin);//freopen(".out", "w", stdout);#endifint T;scanf("%d", &T);while(T--) {scanf("%d", &m);tot = 0;memset(od, 0, sizeof(od));memset(id, 0, sizeof(id));memset(vis, 0, sizeof(vis));memset(head, -1, sizeof(head));char s[1005];for(int i = 0; i  < m; i++) {scanf("%s", s);int len = strlen(s);int u = s[0] - 'a', v = s[len-1] - 'a';od[u]++; id[v]++;vis[u] = vis[v] = 1;add(u, v);}bool flag = Euler() && connect();if(flag) puts("Ordering is possible.");else puts("The door cannot be opened.");}return 0;}