UVA10129 Play on Words —— 欧拉回路

// UVa10129 Play on Words
// Rujia Liu
// 题意：输入n个单词，是否可以排成一个序列，使得每个单词的第一个字母和上一个单词的最后一个字母相同
// 算法：把字母看作结点，单词看成有向边，则有解当且仅当图中有欧拉路径。注意要先判连通
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int maxn = 10000 + 5;

int pa[256], used[256], deg[256]; // 是否出现过；度数
char word[maxn];
int n, cc;

int findset(int x) { return ( pa[x] != x ? pa[x] = findset(pa[x]) : x ); }

void init()
{
memset(used, 0, sizeof(used));
memset(deg, 0, sizeof(deg));

for(int ch = 'a'; ch <= 'z'; ch++)
pa[ch] = ch; // 初始化并查集

cc = 26; // 连通块个数
}

void solve()
{
scanf("%d", &n);
for(int i = 0; i < n; i++)
{
scanf("%s", word);
char c1 = word[0];
char c2 = word[strlen(word)-1];
deg[c1]++;
deg[c2]--;
used[c1] = used[c2] = 1;

int s1 = findset(c1);
int s2 = findset(c2);
if(s1 != s2)
{
pa[s1] = s2;
cc--;
}
}

vector<int> d;
for(int ch = 'a'; ch <= 'z'; ch++)
{
if(!used[ch])
cc--; // 没出现过的字母

else if(deg[ch] != 0)
d.push_back(deg[ch]);
}

int  ok = false;
if(cc == 1 && (d.empty() || (d.size() == 2 && (d[0] == 1 || d[0] == -1))))
ok = true;

if(ok)
printf("Ordering is possible.\n");
else
printf("The door cannot be opened.\n");
}

int main()
{
int T;
scanf("%d", &T);
while(T--)
{
init();
solve();
}
return 0;
}