UVA10129 Play On Words 题解

problem

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 N that indicates the number of plates (1 ≤ N ≤ 100000). Then exactly N lines 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

3
2
acm
ibm
3
acm
malform
mouse
2
ok
ok

Sample Output

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

中文释义

给定一串单词，能否找到一个序列使单词首尾相连，相连的规则是: 上一个单词的尾部字符等于下一个单词的首部字符, 其中输入的单词可能重复

Solution

• 暴力DFS && 枚举 会 TLE
• 每个单词的有效信息实际上只有首尾字符, 不需要存取整个单词

思路一

暴力DFS整张图, 统计每次DFS边的个数是否等于单词数

每个单词是一条边,单词的首尾字符是两个节点,建图

• 会TLE
• 输入包含重复 单词&&边,处理很麻烦

思路二

问题的核心是找到一条经过每条边且仅仅经过一次的路径, 在图论中,这条路径无疑是欧拉道路&&欧拉回路,那么只需判断是否存在欧拉道路即可

欧拉道路 && 欧拉回路

即图论中的七桥问题和一笔画,维基百科中有简略介绍

图类型	欧拉道路	欧拉回路
无向图	只包含两个奇度数节点	不包含奇度数节点
有向图	基图联通 && 最多两个点出度!=入度 &&一个点 入度-出度=1 && 另一个点 出度- 入度=1	省略

所以只需判断两个条件,问题就迎刃而解

• 基图联通

  判断基图联通可以用DFS 或 并查集 判断

• 节点度数符合上述条件

本题很明显是有向图

代码

#include
#include
using namespace std;
const int maxn = 1000 +10;
int n,m;
int point[30], side[30][30],in[30],out[30]. vis[30];
bool euler();

void dfs(int);
//dfs后判断是否有节点未被访问，如果有则图不联通
bool check()
{
     
    for(int i=0;i<=25;i++)
        if(point[i] )
            if(vis[i]==0)
                return false;
    return true;
}

int main()
{
     
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
     
        scanf("%d",&m);
        char buffer[maxn];
        int b;
        for(int i=1;i<=m;i++)
        {
     
            scanf("%s",buffer);
            int a=buffer[0]-'a';
            b= buffer[strlen(buffer)-1]-'a';
            point[a]=1;
            point[b] =1;
            side[a][b]=side[b][a] =1; //判断的是基图联通，有向图的基图即无向图
            in[a]++;//统计入度
            out[b]++;//统计出度
        }
        dfs(b);//无向图的联通就随便找个点遍历，
        if((euler() && check())|| m==1 || m==0) //特判空输入情况
        {
     
            printf("Ordering is possible.\n");
        }
        else printf("The door cannot be opened.\n");
        //清除中间变量
        memset(vis,0,sizeof(vis));
        memset(side,0,sizeof(side));
        memset(point,0,sizeof(point));
        memset(in,0,sizeof(in)); 
        memset(out,0,sizeof(out)); 
    } 
    return 0; 
} 
void dfs(int now)
{
     
    vis[now]=1;
    for(int i=0;i<=25;i++)
        if(side[now][i] && !vis[i]) 
            dfs(i);
}
bool euler() //判断节点度数
{
     
    int count1=0;
    int count2=0;
    for(int i=0;i<=25;i++)
    {
     
        if(in[i]-out[i]>1 && out[i] - in[i]>1)
            return false;
        if(in[i]-out[i]==1)
        {
     
            if(++count1>1) return false;
        }else if(out[i]-in[i]==1)
        {
     
            if(++count2>1)   return false;
        }
    }
    return true;
}

输出格式

本题不像过往的UVA输出格式, 输出文件最后一行必须是空行,注意不要被卡格式,同时对于不包含任意单词的情况需要特判

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