「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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