D-Word

思路 ：

• 一共最多只有2000个字符串，字符串的最大长度才20，而且只有一个询问
• 如果两个字符串之间只有一个或者没有字母不同说明它们之间有一条边；因此，转换成了最短路问题，附加打印路径

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define endl '\n'
#define _(a) cout << #a << ": " << (a) << "    "
#define one first
#define two second
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;
const int N = 2e3 + 10;

int n, m;
string a[N];
vector<int> g[N];
int d[N];
bool st[N];
int fa[N];

bool check(string u, string v) {
    int diff = 0;
    for (int i = 0; i < m; ++ i) {
        diff += (u[i] != v[i]);
    }
    return (diff <= 1);
}
void dijkstra() {
    memset(d, 0x3f, sizeof d);
    priority_queue<pii, vector<pii>, greater<pii>> pq;
    pq.push({0, n + 1});
    d[n + 1] = 0;
    while (pq.size()) {
        pii t = pq.top(); pq.pop();
        int u = t.two;
        if (st[u]) continue;
        st[u] = true;
        for (int v : g[u]) {
            if (d[v] > d[u] + 1) {
                fa[v] = u;
                d[v] = d[u] + 1;
                pq.push({d[v], v});
            }
        }
    }
}

int main() {
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n + 2; ++ i) cin >> a[i];
    for (int i = 1; i <= n + 2; ++ i) {
        for (int j = i + 1; j <= n + 2; ++ j) {
            if (check(a[i], a[j])) {
                g[i].pb(j);
                g[j].pb(i);
            }
        }
    }
    dijkstra();
    if (d[n + 2] == 0x3f3f3f3f) cout << -1;
    else {
        cout << d[n + 2] - 1 << endl;
        vector<int> ans;
        int now = n + 2;
        while (now != n + 1) {
            ans.push_back(now);
            now = fa[now];
        }
        ans.push_back(n + 1);
        for (int i = (int)ans.size() - 1; i >= 0; -- i) cout << a[ans[i]] << endl;
    }
}

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