HDU 1386--Play on Words【并查集判连通 + 欧拉路】

Play on Words

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 10589 Accepted: 3596

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.

判断欧拉路是否存在的方法
有向图：图连通，有一个顶点出度大入度1，有一个顶点入度大出度1，其余都是出度=入度。
无向图：图连通，只有两个顶点是奇数度，其余都是偶数度的。
判断欧拉回路是否存在的方法
有向图：图连通，所有的顶点出度=入度。
无向图：图连通，所有顶点都是偶数度。

思路：每组单词可以构造一个图，图中定点为26个小写字母， 每个单词为图中的一条边，而且图是有向图。

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <cstdlib>#define maxn 110000using namespace std;int per[30], in[30], out[30], vis[30];int M;void init(){    for(int i = 0; i < 26; ++i)        per[i] = i;    memset(in, 0, sizeof(in));    memset(out, 0 ,sizeof(out));    memset(vis, 0 ,sizeof(vis));}int find (int x){    int r = x;    while(r != per[r])        r = per[r];    per[x] = r;    return r;}void join (int a, int b){    int fa = find(a);    int fb = find(b);    if(fa != fb)        per[fa] = fb;}int main(){    int T;    char str[1100];    scanf("%d", &T);    while(T--){        scanf("%d", &M);        init();        for(int i = 0; i < M; ++i){            scanf("%s", str);            int len, u, v;            len = strlen(str);// u --> v            u = str[0] - 'a';            v = str[len - 1] - 'a';            out[u]++, in[v]++;            vis[u] = vis[v] = 1;            join(u, v);        }        int ans = 0;        for(int i = 0; i <= 25; ++i){            if(per[i] == i && vis[i])                ans++;            if(ans > 1)                break;        }        if(ans > 1){            printf("The door cannot be opened.\n");// 没有构成通路            continue;        }        int st, ed;        st = ed = 0;        int flag = 1;        for(int i = 0; i <26; ++i){            if(vis[i] && in[i] != out[i]){                if(in[i] - out[i] == 1)                    st++;                else if(out[i] - in[i] == 1)                    ed++;                else{                    flag = 0;                    break;                }            }        }        //printf("st = %d, ed = %d\n", st, ed);        if(flag && st <= 1 && ed <= 1)//st == 0 && ed == 0 是考虑到只有单词只有 aa的这种情况            printf("Ordering is possible.\n");        else            printf("The door cannot be opened.\n");    }    return 0;}