[CF508D] Tanya and Password - 欧拉通路,hash
Description
有一个长度为 (n+2) 的字符串 (S),现在给你 (S[1..3],S[2..4],...,S[n,n+2]),但是是打乱顺序的,你需要构造出 (S)
Solution
将每个给定的串 (s[1..3]) 的转化为图上的一条边 (H(s[1],s[2]) o H(s[2],s[3])),然后跑欧拉通路即可
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int M = 128;
const int N = 100005;
int n;
vector<int> g[N];
vector<int> ans;
int din[N], dout[N];
void dfs(int p)
{
while (g[p].size())
{
int q = g[p].back();
g[p].pop_back();
dfs(q);
}
ans.push_back(p);
}
signed main()
{
ios::sync_with_stdio(false);
cin >> n;
for (int i = 1; i <= n; i++)
{
string s;
cin >> s;
g[s[0] * M + s[1]].push_back(s[1] * M + s[2]);
dout[s[0] * M + s[1]]++;
din[s[1] * M + s[2]]++;
}
map<int, int> mp;
for (int i = 1; i <= M * M; i++)
mp[din[i] - dout[i]]++;
if (!((mp.size() == 1) || (mp.size() == 3 && mp[1] == 1 && mp[-1] == 1)))
{
cout << "NO" << endl;
}
else
{
int start = 0;
for (int i = 1; i <= M * M; i++)
if (dout[i])
start = i;
for (int i = 1; i <= M * M; i++)
if (dout[i] - din[i] == 1)
start = i;
dfs(start);
if (ans.size() != n + 1)
{
cout << "NO" << endl;
return 0;
}
cout << "YES" << endl;
for (int i = ans.size() - 1; i >= 0; i--)
cout << (char)(ans[i] / M);
cout << (char)(ans[0] % M) << endl;
}
}