OI
8 minutes read (About 1163 words)

「模拟测试」20171030

T2 Game

甲乙两个人轮流那一些物品，甲先手，他可以拿走 $1$ 或 $2$ 个物品。对于后面，若前一个人拿走 $k$ 个物品，当前的人可以拿走 $k$ 或 $k + 1$ 个物品，甲乙的策略都是让自己尽量比别人拿的物品的价值高，求最优策略下，甲最多比乙多拿多少？

题解

$f[i][j]$ 表示第 $i$ 个物品，当前拿 $j$ 个时，甲最多比乙拿多少，先预处理一个前缀和。
注意到其实甲的决策并不会影响到答案，只要乙的决策最劣即是最优答案，而且显然 $j \leq \sqrt{n}$，所以我们直接记忆化搜索就好了。

代码

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/**
 * Copyright (c) 2017, xehoth
 * All rights reserved.
 * 「SuperOJ 2018」游戏 30-10-2017
 * DP
 * @author xehoth
 */
#include <bits/stdc++.h>

namespace IO {

inline char read() {
    static const int IN_LEN = 1000000;
    static char buf[IN_LEN], *s, *t;
    s == t ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin) : 0;
    return s == t ? -1 : *s++;
}

template <typename T>
inline bool read(T &x) {
    static char c;
    static bool iosig;
    for (c = read(), iosig = false; !isdigit(c); c = read()) {
        if (c == -1) return false;
        c == '-' ? iosig = true : 0;
    }
    for (x = 0; isdigit(c); c = read()) x = x * 10 + (c ^ '0');
    iosig ? x = -x : 0;
    return true;
}

inline void read(char &c) {
    while (c = read(), isspace(c) && c != -1)
        ;
}

inline int read(char *buf) {
    register int s = 0;
    register char c;
    while (c = read(), isspace(c) && c != -1)
        ;
    if (c == -1) {
        *buf = 0;
        return -1;
    }
    do
        buf[s++] = c;
    while (c = read(), !isspace(c) && c != -1);
    buf[s] = 0;
    return s;
}

const int OUT_LEN = 1000000;

char obuf[OUT_LEN], *oh = obuf;

inline void print(char c) {
    oh == obuf + OUT_LEN ? (fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf) : 0;
    *oh++ = c;
}

template <typename T>
inline void print(T x) {
    static int buf[30], cnt;
    if (x == 0) {
        print('0');
    } else {
        x < 0 ? (print('-'), x = -x) : 0;
        for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 | 48;
        while (cnt) print((char)buf[cnt--]);
    }
}

inline void print(const char *s) {
    for (; *s; s++) print(*s);
}

inline void flush() { fwrite(obuf, 1, oh - obuf, stdout); }

struct InputOutputStream {
    template <typename T>
    inline InputOutputStream &operator>>(T &x) {
        read(x);
        return *this;
    }

    template <typename T>
    inline InputOutputStream &operator<<(const T &x) {
        print(x);
        return *this;
    }

    ~InputOutputStream() { flush(); }
} io;
}

namespace {

const int MAXN = 20000;
const int INF = -1061109568;

using IO::io;

int a[MAXN + 1], sum[MAXN + 1], f[MAXN + 1][255], n;

inline int optMax(const register int a, const register int b) {
    return a > b ? a : b;
}

inline int dp(register int x, register int k) {
    register int &ret = f[x][k];
    if (ret != INF) return ret;
    return x + k > n
               ? 0
               : (x + k <= n
                      ? ret = optMax(ret, sum[x + k] - sum[x] - dp(x + k, k))
                      : 0,
                  x + k + 1 <= n
                      ? ret = optMax(
                            ret, sum[x + k + 1] - sum[x] - dp(x + k + 1, k + 1))
                      : 0,
                  ret);
}

inline void solve() {
    register int T;
    io >> T;
    while (T--) {
        io >> n;
        for (register int i = 1; i <= n; i++)
            io >> a[i], sum[i] = sum[i - 1] + a[i];
        memset(f, 0xc0, sizeof(f));
        io << dp(0, 1) << '\n';
    }
}
}

int main() {
    // freopen("sample/1.in", "r", stdin);
    solve();
    return 0;
}

T3 星星

给出一个无重边自环的无向图，求由四个点五条变组成的这种图的个数
T3 星星

题解

对于每个点我们先把它连的点打上标记，再枚举所有与它相连且度数比它小的点的边集来求每条边的贡献，可以证明这样的复杂度是 $O(m \sqrt{m})$ 的。

至于证明我们分度数 $\sqrt{m}$ 来讨论即可。

代码

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/**
 * Copyright (c) 2017, xehoth
 * All rights reserved.
 * 「SuperOJ 2019」星星 30-10-2017
 *
 * @author xehoth
 */
#include <bits/stdc++.h>

namespace IO {

inline char read() {
    static const int IN_LEN = 1000000;
    static char buf[IN_LEN], *s, *t;
    s == t ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin) : 0;
    return s == t ? -1 : *s++;
}

template <typename T>
inline bool read(T &x) {
    static char c;
    static bool iosig;
    for (c = read(), iosig = false; !isdigit(c); c = read()) {
        if (c == -1) return false;
        c == '-' ? iosig = true : 0;
    }
    for (x = 0; isdigit(c); c = read()) x = x * 10 + (c ^ '0');
    iosig ? x = -x : 0;
    return true;
}

inline void read(char &c) {
    while (c = read(), isspace(c) && c != -1)
        ;
}

inline int read(char *buf) {
    register int s = 0;
    register char c;
    while (c = read(), isspace(c) && c != -1)
        ;
    if (c == -1) {
        *buf = 0;
        return -1;
    }
    do
        buf[s++] = c;
    while (c = read(), !isspace(c) && c != -1);
    buf[s] = 0;
    return s;
}

const int OUT_LEN = 1000000;

char obuf[OUT_LEN], *oh = obuf;

inline void print(char c) {
    oh == obuf + OUT_LEN ? (fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf) : 0;
    *oh++ = c;
}

template <typename T>
inline void print(T x) {
    static int buf[30], cnt;
    if (x == 0) {
        print('0');
    } else {
        x < 0 ? (print('-'), x = -x) : 0;
        for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 | 48;
        while (cnt) print((char)buf[cnt--]);
    }
}

inline void print(const char *s) {
    for (; *s; s++) print(*s);
}

inline void flush() { fwrite(obuf, 1, oh - obuf, stdout); }

struct InputOutputStream {
    template <typename T>
    inline InputOutputStream &operator>>(T &x) {
        read(x);
        return *this;
    }

    template <typename T>
    inline InputOutputStream &operator<<(const T &x) {
        print(x);
        return *this;
    }

    ~InputOutputStream() { flush(); }
} io;
}

namespace {

using IO::io;

const int MAXN = 100000;
int n, m;

struct Data {
    int deg, id;
    inline bool operator<(const Data &d) const { return deg > d.deg; }
} d[MAXN + 1];

std::vector<int> edge[MAXN + 1];

typedef std::vector<int>::iterator Iterator;
bool vis[MAXN + 1];
int tag[MAXN + 1];

inline void solveCase() {
    for (register int i = 1, u, v; i <= m; i++) {
        io >> u >> v, edge[u].push_back(v), edge[v].push_back(u);
        d[u].deg++, d[v].deg++;
    }
    for (register int i = 1; i <= n; i++) d[i].id = i;
    memset(vis, false, sizeof(bool) * (n + 1));
    memset(tag, 0, sizeof(int) * (n + 1));
    register long long ans = 0;
    std::sort(d + 1, d + n + 1);
    for (register int i = 1, cur, cnt; i <= n; i++) {
        vis[cur = d[i].id] = true;
        for (register Iterator p = edge[cur].begin(); p != edge[cur].end(); p++)
            tag[*p] = cur;
        for (register Iterator p = edge[cur].begin(); p != edge[cur].end();
             p++) {
            cnt = 0;
            if (!vis[*p])
                for (register Iterator k = edge[*p].begin();
                     k != edge[*p].end(); k++)
                    if (tag[*k] == cur) cnt++;
            ans += ((long long)cnt * (long long)(cnt - 1)) / 2;
        }
    }
    io << ans << '\n';
    for (register int i = 1; i <= n; i++) edge[i].clear();
    for (register int i = 1; i <= n; i++) d[i].deg = 0;
}

inline void solve() {
    register int T;
    for (io >> T, T--; T--;) io >> n >> m, solveCase();
    io >> n >> m, solveCase();
}
}

int main() {
    // freopen("sample/1.in", "r", stdin);
    solve();
    return 0;
}
# ,
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