Luogu 1341 无序字母对

Solution

找一条字典序最小的欧拉路径。

用 $multiset$ 存储领接表。

欧拉路径模板传送门

Code

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
using namespace std;

const int N = 1e3 + 5;

int n, up = 'z' - 'A' + 1;
int ans[N], tot, d[N], num;

multiset<int> to[N];

void dfs(int u) {
    multiset<int> :: iterator i;
    for (i = to[u].begin(); i != to[u].end(); i = to[u].begin()) {
        int nt = *i;
        to[nt].erase(to[nt].find(u));
        to[u].erase(to[u].find(nt));
        dfs(nt);
    }
    ans[++tot] = u;
}

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        char s[3];
        scanf("%s", s);
        to[s[0] - 'A' + 1].insert(s[1] - 'A' + 1);
        to[s[1] - 'A' + 1].insert(s[0] - 'A' + 1);
        d[s[1] - 'A' + 1]++;
        d[s[0] - 'A' + 1]++;
    }
    for (int i = 1; i <= up; ++i)
        if (d[i] & 1)
            num++;
    if (num == 1 || num > 2)
        return puts("No Solution"), 0;
    if (num == 0) {
        for (int i = 1; i <= up; ++i)
            if (d[i]) {dfs(i); break;}
    }
    else 
        for (int i = 1; i <= up; ++i)
            if (d[i] && (d[i] & 1)) {dfs(i); break;}
    for (int i = tot; i; i--) 
        putchar(ans[i] + 'A' - 1);
    puts("");
}