POJ 1386 Play on Words（欧拉图的判断）

Play on Words

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11838		Accepted: 4048

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

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.

 

 

题目链接：POJ 1386

做法：把单词看成有向边，比如acm就是从a->m的一条边，但是字母肯定不能当节点啊，由于字母都是lowercase，那就都减掉'a'来映射到数字0~25就可以了（单词的长度其实在这个题里显然没什么卵用）然后进行欧拉图判断

对于是否是欧拉图的判断：首先基图要连通，然后再看入度和出度不相等的点的个数，若为0，则为欧拉图且有欧拉回路；如果为1，即只有一个点入度不等于其他点均相等，则出度不会等于入度，显然这在图里是不可能的，图的入度和一定会等于出度和；然后是两个点，看是否一个是入度比出度少1，一个是入度比出度多1……，忘记判断vector为空RE几次…………

代码：

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=100010;
const int L=1005;
char s[L];
struct edge
{
    int to;
    int pre;
};
edge E[N<<1];
int head[N],tot;
int cur[30],vis[30];
int ru[30],chu[30];
void init()
{
    CLR(head,-1);
    tot=0;
    CLR(vis,0);
    CLR(cur,0);
    CLR(ru,0);
    CLR(chu,0);
}
void add(int s,int t)
{
    E[tot].to=t;
    E[tot].pre=head[s];
    head[s]=tot++;
}
bool bfs(int s,int all)
{
    queue<int>Q;
    vis[s]=1;
    --all;
    Q.push(s);
    while (!Q.empty())
    {
        int now=Q.front();
        Q.pop();
        for (int i=head[now]; ~i; i=E[i].pre)
        {
            int v=E[i].to;
            if(!vis[v])
            {
                vis[v]=1;
                --all;
                Q.push(v);
            }
        }
    }
    return !all;
}
int main(void)
{
    int tcase,n,i,a,b,len;
    scanf("%d",&tcase);
    while (tcase--)
    {
        init();
        scanf("%d",&n);
        int P=0;
        while (n--)
        {
            scanf("%s",s);
            len=strlen(s);
            a=s[0]-'a';
            b=s[len-1]-'a';
            if(!cur[a])
                cur[a]=1,++P;
            if(!cur[b])
                cur[b]=1,++P;
            add(a,b);
            add(b,a);
            ++chu[a];
            ++ru[b];
        }
        int s=0;
        for (i=0; i<26; ++i)
        {
            if(cur[i])
            {
                s=i;
                break;
            }
        }
        bool flag=bfs(s,P);
        if(!flag)
            puts("The door cannot be opened.");
        else
        {
            vector<pii>pos;
            for (i=0; i<26; ++i)
            {
                if((ru[i]==0&&chu[i]==0)||(ru[i]==chu[i]))
                    continue;
                pos.push_back(pii(ru[i],chu[i]));
            }
            if(pos.empty())
                flag=true;
            else if(pos.size()>2)
                flag=false;
            else
            {
                if(pos[0].first==pos[0].second+1&&pos[1].first==pos[1].second-1)
                    flag=true;
                else if(pos[1].first==pos[1].second+1&&pos[0].first==pos[0].second-1)
                    flag=true;
                else
                    flag=false;
            }
            puts(flag?"Ordering is possible.":"The door cannot be opened.");          
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/Blackops/p/5897527.html