Wow! Such String!

標籤：acm

題目連結

• 題意：
  給一個n，輸出一個長度為n的字串，使得串中任意相連四個字元組成的串不重複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:歐拉迴路  返回2：歐拉路經    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;}

其實，如果沒有想到是歐拉迴路問題，暴力也是可以試試的，懊悔當時沒寫出來，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;}