20171109校内模拟赛

 

 

这场考试完美的展示了lqz高超的智力水平

竟然能够直逼noip胎教组!

 

 

应得:100 + 100 + 100

实得:40   + 100 + 100

 

 

T1简直zz,一开始一看题,woc,这么裸!?于是就tarjan,缩点,拓扑balabala

突然一想,这直接dfs不就好了吗?然后就丢掉了刚才的代码,转去写dfs

写完一边过了样例就丢到一边不管了。。。。然后就40分

 

突然意识到。。。mdzz,应该是建反向边,我建的正向边。。。智商感人。

 

T1

 

/*
    zz的dfs
40分
*/ #include <cstdio> #include <iostream> #define rg register inline void read (int &n) { rg char c = getchar (); bool temp = false; for (n = 0; !isdigit (c); c = getchar ()) if (c == '-') temp = true; for (; isdigit (c); n = n * 10 + c - '0', c = getchar ()); if (temp) n = -n; } #define Max 300009 bool is[Max]; struct E { E *n; int v; } *list[Max], pool[Max], *Ta = pool; inline void In (int u, int v) { ++ Ta, Ta->v = v, Ta->n = list[u], list[u] = Ta; } int f[Max]; inline void cmax (int &a, int b) { if (b > a) a = b; } inline int max (int a, int b) { return a > b ? a : b; } void Dfs (int n) { cmax (f[n], n); rg int v; for (E *e = list[n]; e; e = e->n) { if (!is[v = e->v]) is[v] = true, Dfs (v); cmax (f[n], max (f[v], v)); } } std :: string Name = "graph", _I = ".in", _O = ".out"; int main (int argc, char *argv[]) { freopen ((Name + _I).c_str (), "r", stdin); freopen ((Name + _O).c_str (), "w", stdout); int x, y, N, M; read (N), read (M); rg int i; for (i = 1; i <= M; ++ i) read (x), read (y), In (x, y); for (i = 1; i <= N; ++ i) if (!is[i]) Dfs (i); for (i = 1; i <= N; ++ i) printf ("%d ", f[i]); return 0; }

 

 

/*
    zz的tajan
90分
*/ #include <cstdio> #include <iostream> #include <stack> #include <queue> #define rg register inline void read (int &n) { rg char c = getchar (); for (n = 0; !isdigit (c); c = getchar ()); for (; isdigit (c); n = n * 10 + c - '0', c = getchar ()); } std :: string Name = "graph", _I = ".in", _O = ".out"; #define Max 200005 struct Graph { int l[Max], v[Max], n[Max], EC; Graph () { EC = 0; } inline void In (int x, int y) { v[++ EC] = y, n[EC] = l[x], l[x] = EC; } } g1, g2; std :: stack <int> S; int dfn[Max], low[Max], scc[Max], SC; int key[Max], N, DC; inline void cmin (int &a, int b) { if (b < a) a = b; } inline void cmax (int &a, int b) { if (b > a) a = b; } void Dfs (int n) { S.push (n), dfn[n] = low[n] = ++ DC; rg int i, v; for (i = g1.l[n]; i; i = g1.n[i]) if (!dfn[v = g1.v[i]]) Dfs (v), cmin (low[n], low[v]); else if (!scc[v]) cmin (low[n], dfn[v]); if (dfn[n] == low[n]) for (v = n + 1, ++ SC; !S.empty () && v != n; S.pop ()) v = S.top (), scc[v] = SC, cmax (key[SC], v); } int in[Max]; int Find (int x, int y) { for (rg int i = g2.l[x]; i; i = g2.n[i]) if (g2.v[i] == y) return true; return false; } void Tarjan () { rg int i; for (i = 1; i <= N; ++ i) if (!dfn[i]) Dfs (i); for (int n = 1; n <= N; ++ n) for (i = g1.l[n]; i; i = g1.n[i]) if (scc[n] != scc[g1.v[i]] && !Find (scc[n], scc[g1.v[i]])) g2.In (scc[n], scc[g1.v[i]]), ++ in[scc[g1.v[i]]]; } int f[Max]; inline int max (int a, int b) { return a > b ? a : b; } void Get (int n) { cmax (f[n], key[n]); for (rg int i = g2.l[n]; i; i = g2.n[i]) { Get (g2.v[i]); cmax (f[n], max (f[g2.v[i]], key[g2.v[i]])); } } void GetAnswer () { rg int i; for (i = 1; i <= SC; ++ i) if (!in[i]) Get (i); for (i = 1; i <= N; ++ i) printf ("%d ", f[scc[i]]); } int main (int argc, char *argv[]) { freopen ((Name + _I).c_str (), "r", stdin); freopen ((Name + _O).c_str (), "w", stdout); int M, x, y; read (N), read (M); rg int i; for (i = 1; i <= M; ++ i) read (x), read (y), g1.In (x, y); Tarjan (); GetAnswer (); return 0; }
/*
    反向边dfs
    100分
*/
#include <cstdio>
#include <iostream>
#define rg register

inline void read (int &n)
{
    rg char c = getchar (); bool temp = false;
    for (n = 0; !isdigit (c); c = getchar ()) if (c == '-') temp = true;
    for (; isdigit (c); n = n * 10 + c - '0', c = getchar ());
    if (temp) n = -n;
}

#define Max 300009
bool is[Max];
struct E { E *n; int v; } *list[Max], pool[Max], *Ta = pool;

inline void In (int u, int v)
{ ++ Ta, Ta->v = v, Ta->n = list[u], list[u] = Ta; }
int f[Max];
inline void cmax (int &a, int b) { if (b > a) a = b; }
inline int max (int a, int b) { return a > b ? a : b; }
void Dfs (int n)
{
    rg int v;
    for (E *e = list[n]; e; e = e->n)
        if (!is[v = e->v]) 
            cmax (f[v], f[n]), is[v] = true, Dfs (v);
}
std :: string Name = "graph", _I = ".in", _O = ".out";
int main (int argc, char *argv[])
{
    freopen ((Name + _I).c_str (), "r", stdin);
    freopen ((Name + _O).c_str (), "w", stdout);
    int x, y, N, M; read (N), read (M); rg int i;
    
    for (i = 1; i <= M; ++ i)
        read (x), read (y), In (y, x);
    
    for (i = 1; i <= N; ++ i) f[i] = i;
    for (i = N; i >= 1; -- i)
        if (!is[i]) Dfs (i);
        
    for (i = 1; i <= N; ++ i) printf ("%d ", f[i]);
    return 0;
}

 

 

T2

 

 

/*
    nlogn的最长严格上升子序列 
*/
#include <cstdio>
#include <iostream>
#include <algorithm>

#define rg register

inline void read (int &n)
{
    rg char c = getchar (); bool temp = false;
    for (n = 0; !isdigit (c); c = getchar ()) if (c == '-') temp = true;
    for (; isdigit (c); n = n * 10 + c - '0', c = getchar ());
    if (temp) n = -n;
}

#define INF 1e9
#define Max 100009

int a[Max], f[Max];
std :: string Name = "incr", _I = ".in", _O = ".out";
inline void cmax (int &a, int b) { if (b > a) a = b; }
int main (int argc, char *argv[])
{
    freopen ((Name + _I).c_str (), "r", stdin);
    freopen ((Name + _O).c_str (), "w", stdout);
    int N, t = 1, p; read (N); rg int i; f[0] = -INF;
    for (i = 1; i <= N; ++ i) read (a[i]);
    f[t] = -- a[t];
    for (i = 2; i <= N; ++ i)
    {
        p = std :: upper_bound (f + 1, f + 1 + t, a[i] - i) - f;
        f[p] = a[i] - i, cmax (t, p);
    }
    printf ("%d", N - t);
    return 0;
}

 

 

 

 

T3

 

/*
    高精+dp 
*/
#include <iostream>
#include <cstdio>
#include <vector>
#include <iomanip>

#define rg register

inline void read (int &n)
{
    rg char c = getchar (); bool temp = false;
    for (n = 0; !isdigit (c); c = getchar ()) if (c == '-') temp = true;
    for (; isdigit (c); n = n * 10 + c - '0', c = getchar ());
    if (temp) n = -n;
}

#define Max 200

const int BB = 1000000000, BL = 9;

struct Int 
{
    inline int max (int a, int b) { return a > b ? a : b; }
    
    std :: vector <int> c; Int () {} typedef long long LL; 
    
    Int (int x) { for (; x > 0; c.push_back (x % BB), x /= BB); }
    inline void CrZ () { for (; !c.empty () && c.back () == 0; c.pop_back ()); }
    
    inline Int &operator += (const Int &rhs)
    {
        c.resize (max (c.size (), rhs.c.size ())); rg int i, t = 0, S;
        for (i = 0, S = rhs.c.size (); i < S; ++ i)
            c[i] += rhs.c[i] + t, t = c[i] >= BB, c[i] -= BB & (-t);
        for (i = rhs.c.size (), S = c.size (); t && i < S; ++ i)
            c[i] += t, t = c[i] >= BB, c[i] -= BB & (-t);
        if (t) c.push_back (t); return *this;
    }
    
    inline Int &operator *= (const Int &rhs)
    {
        rg int na = c.size (), i, j, S, ai; 
        c.resize (na + rhs.c.size ()); LL t;
        for (i = na - 1; i >= 0; -- i)
        {
            ai = c[i], t = 0, c[i] = 0;
            for (j = 0, S = rhs.c.size (); j < S; ++ j)
                t += c[i + j] + (LL) ai * rhs.c[j], c[i + j] = t % BB, t /= BB;
            for (j = rhs.c.size (), S = c.size (); t != 0 && i + j < S; ++ j)
                t += c[i + j], c[i + j] = t % BB, t /= BB;
        }
        CrZ (); return *this;
    }
    
    friend inline Int operator + (const Int &lhs, const Int &rhs)
    { Int res = lhs; return res += rhs; }
    friend inline Int operator * (const Int &lhs, const Int &rhs)
    { Int res = lhs; return res *= rhs; }
    
    friend inline std :: ostream &operator << (std :: ostream &out, const Int &rhs)
    { 
        if (rhs.c.size () == 0) out << "0";
        else 
        {
            out << rhs.c.back ();
            for (rg int i = rhs.c.size () - 2; i >= 0; -- i)
                out << std :: setfill ('0') << std :: setw (BL) << rhs.c[i];
        }
        return out;
    } 
};

Int f[Max][Max];
std :: string Name = "permutation", _I = ".in", _O = ".out";

int main (int argc, char *argv[])
{
    freopen ((Name + _I).c_str (), "r", stdin);
    freopen ((Name + _O).c_str (), "w", stdout);
    int N, K; read (N), read (K); rg int i, j;
    if (K >= N) return puts ("0"), 0;
    
    for (i = 1; i <= N; ++ i) f[i][0] = 1;
    
    for (i = 2; i <= N; ++ i)
        for (j = 1; j <= K; ++ j)
            if (j < i) f[i][j] += f[i - 1][j - 1] * (i - j) + (j + 1) * f[i - 1][j];
                
    std :: cout << f[N][K];
    return 0;
}

 

posted @ 2017-11-09 15:00  ZlycerQan  阅读(198)  评论(0编辑  收藏  举报