首页 > 代码库 > Wow! Such String!

Wow! Such String!

2024-07-16 01:08:29   222人阅读

题目链接

  • 题意：
    给一个n，输出一个长度为n的字符串，使得串中任意相连四个字符组成的串不重复N (1 ≤ N ≤ 500000)
  • 分析：
const int MAXV = 18278;
const int MAXE = 475228;

struct Edge
{
    int from, to;
};

struct Direct_Euler
{
    int n, m;
    int out[MAXV];
    bool vis[MAXE];
    vector<int> G[MAXV];
    vector<Edge> edges;
    stack<int> sk;
    vector<int> ans;

    void init(int n)
    {
        this->n = n;
        edges.clear();
        REP(i, n)
        {
            out[i] = 0;
            G[i].clear();
        }
    }

    void addEdge(int a, int b)
    {
        edges.push_back((Edge) { a, b });
        m = edges.size();
        G[a].push_back(m - 1);
        out[a]++; out[b]--;
    }

    void dfs(int u, int ind)
    {
        REP(i, G[u].size())
        {
            int x = G[u][i];
            Edge& e = edges[x];
            if (!vis[x])
            {
                vis[x] = true;
                dfs(e.to, x);
            }
        }
        if (ind >= 0)
            sk.push(ind);
    }

    //返回1:欧拉回路  返回2：欧拉路经
    int solve(int s)
    {
        int cnt = 0;
        ans.clear();
        REP(i, n)
        {
            if (out[i] == 1)
            {
                if (++cnt > 1)
                {
                    return 0;
                }
                s = i;
            }
            else if (out[i] > 1 || out[i] < -1)
                return 0;
        }
        while (!sk.empty()) sk.pop();
        REP(i, m) vis[i] = false;
        dfs(s, -1);
        REP(i, m)
            if (!vis[i])
                return 0;
        while (!sk.empty())
        {
            ans.push_back(sk.top());
            sk.pop();
        }
        return cnt != 0 ? 1 : 2;
    }
} graph;
char x[456979], tot = 0;
int main()
{
    int size = 29 << 20; // 29MB
    char *p = (char*)malloc(size) + size;
    __asm__("movl %0, %%esp\n" :: "r"(p));

    graph.init(18278);
    REP(i, 26) REP(j, 26) REP(k, 26) REP(l, 26)
        graph.addEdge(i * 676 + j * 26 + k, j * 676 + k * 26 + l);
    graph.solve(0);
    vector<int>& ans = graph.ans;
    int tot = 0;

    int to = graph.edges[ans[0]].from;
    x[tot++] = to / 676 + 'a';
    to %= 676;
    x[tot++] = to / 26 + 'a';
    to %= 26;
    x[tot++] = to + 'a';
    REP(i, ans.size())
        x[tot++] = graph.edges[ans[i]].to % 26 + 'a';
    int n;
    while (~RI(n))
    {
        if (n > tot)
            puts("Impossible");
        else
        {
            REP(i, n)
                printf("%c", x[i]);
            puts("");
        }
    }
    return 0;
}


其实，如果没有想到是欧拉回路问题，暴力也是可以试试的，懊悔当时没写出来，祭奠一下。。
const int MAXN = 1000000;

bool vis[26][26][26][26];
int x[MAXN];
int tot;

int main()
{
    tot = 0;
    REP(i, 26)
    {
        x[tot + 3] = x[tot + 2] = x[tot + 1] = x[tot] = i;
        tot += 4;
    }
    FF(i, 3, tot)
        vis[x[i]][x[i - 1]][x[i - 2]][x[i - 3]] = 1;
    bool flag = true;
    while (flag)
    {
        flag = false;
        REP(i, 26)
        {
            if (!vis[i][x[tot - 1]][x[tot - 2]][x[tot - 3]])
            {
                vis[i][x[tot - 1]][x[tot - 2]][x[tot - 3]] = 1;
                x[tot++] = i;
                flag = true;
            }
        }
    }

    int n;
    while (~RI(n))
    {
        if (n > tot)
            puts("Impossible");
        else
        {
            REP(i, n)
                printf("%c", x[i] + 'a');
            puts("");
        }
    }
    return 0;
}


联系
我们