Codeforces Round #812 (Div. 2)
咕。
B. Optimal Reduction
观察可得：对于所有\(b\)，其中要求操作次数最少的为\(\max \{ a \}\).
然后就是看\(f(a)\)是不是等于\(\max \{ a \}\)了，这个只需要满足\(a\)是单峰的就行。
AC代码
// Problem: B. Optimal Reduction
// Contest: Codeforces - Codeforces Round #812 (Div. 2)
// URL: https://codeforces.com/contest/1713/problem/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define CPPIO std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
#define freep(p) p ? delete p, p = nullptr, void(1) : void(0)
#ifdef BACKLIGHT
#include "debug.h"
#else
#define logd(...) ;
#endif
using i64 = int64_t;
using u64 = uint64_t;
void solve_case(int Case);
int main(int argc, char* argv[]) {
CPPIO;
int T = 1;
std::cin >> T;
for (int t = 1; t <= T; ++t) {
solve_case(t);
}
return 0;
}
void solve_case(int Case) {
int n;
std::cin >> n;
std::vector<int> a(n + 1);
for (int i = 1; i <= n; ++i)
std::cin >> a[i];
std::vector<int> pma(n + 1);
for (int i = 1; i <= n; ++i) {
pma[i] = std::max(pma[i - 1], a[i]);
}
std::vector<int> sma(n + 2);
for (int i = n; i >= 1; --i) {
sma[i] = std::max(sma[i + 1], a[i]);
}
bool flag = true;
for (int i = 2; i <= n - 1; ++i) {
if (a[i] < pma[i - 1] && a[i] < sma[i + 1]) {
flag = false;
}
}
std::cout << (flag ? "YES" : "NO") << "\n";
}
C. Build Permutation
手推了一下，然后猜结论：从n开始往前搞，尽可能构造大的平方数，一定能够得出答案。
AC代码
// Problem: C. Build Permutation
// Contest: Codeforces - Codeforces Round #812 (Div. 2)
// URL: https://codeforces.com/contest/1713/problem/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define CPPIO std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
#define freep(p) p ? delete p, p = nullptr, void(1) : void(0)
#ifdef BACKLIGHT
#include "debug.h"
#else
#define logd(...) ;
#endif
using i64 = int64_t;
using u64 = uint64_t;
void solve_case(int Case);
int main(int argc, char* argv[]) {
CPPIO;
int T = 1;
std::cin >> T;
for (int t = 1; t <= T; ++t) {
solve_case(t);
}
return 0;
}
void solve_case(int Case) {
int n;
std::cin >> n;
std::set<int> s;
for (int i = 0; i < n; ++i) {
s.insert(i);
}
int x = 0;
while ((x + 1) * (x + 1) <= n + n - 1) {
x = x + 1;
}
logd(x, n);
std::vector<int> p(n, -1);
for (int i = n - 1; i >= 0; --i) {
while (true) {
if (x < 0) {
std::cout << "-1\n";
return;
}
int y = x * x - i;
if (s.count(y)) {
p[i] = y;
s.erase(y);
break;
} else {
x = x - 1;
}
}
}
// for (int i = 0; i < n; ++i) {
// int z = int(sqrt(i + p[i]));
// assert(z * z == i + p[i]);
// }
for (int i = 0; i < n; ++i)
std::cout << p[i] << " \n"[i + 1 == n];
}
D. Tournament Countdown
首先，直接朴素的模拟要\(2^n - 1\)次操作，不可行。
然后，观察发现对于\(4\)个人的比赛，其实只需要2次操作就可以确定胜者，然后就可以自底向上4个为一组搞。这样的话，相当于可以用2次操作干掉3个人，一共要干掉\(2^n - 1\)个人，次数就是\(\lceil (2^n - 1)\frac{2}{3} \rceil = \lfloor \frac{2^{n + 1}}{3} \rfloor\)
AC代码
// Problem: D. Tournament Countdown
// Contest: Codeforces - Codeforces Round #812 (Div. 2)
// URL: https://codeforces.com/contest/1713/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define CPPIO std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
#define freep(p) p ? delete p, p = nullptr, void(1) : void(0)
#ifdef BACKLIGHT
#include "debug.h"
#else
#define logd(...) ;
#endif
using i64 = int64_t;
using u64 = uint64_t;
void solve_case(int Case);
int main(int argc, char* argv[]) {
CPPIO;
int T = 1;
std::cin >> T;
for (int t = 1; t <= T; ++t) {
solve_case(t);
}
return 0;
}
std::mt19937 rng(114514);
void solve_case(int Case) {
int n;
#ifdef BACKLIGHT
n = 17;
#else
std::cin >> n;
#endif
int limit = ((1 << (n + 1)) + 2) / 3;
#ifdef BACKLIGHT
std::vector<int> c(1 << n, 0);
std::vector<int> p(1 << n);
std::iota(p.begin(), p.end(), 0);
while (p.size() > 1) {
std::vector<int> np;
for (int i = 0; i < p.size(); i += 2) {
if (rng() & 1) {
np.push_back(p[i + 1]);
++c[p[i + 1]];
} else {
np.push_back(p[i]);
++c[p[i]];
}
}
p = np;
}
int std = p[0];
int count = 0;
auto Q = [&c, &count, &limit](int a, int b) -> int {
++count;
assert(count <= limit);
int r;
if (c[a] == c[b])
r = 0;
else if (c[a] > c[b])
r = 1;
else
r = 2;
return r;
};
#else
int count = 0;
auto Q = [&count, &limit](int a, int b) -> int {
++count;
assert(count <= limit);
std::cout << "? " << a + 1 << " " << b + 1 << std::endl;
int r;
std::cin >> r;
return r;
};
#endif
std::vector<int> a(1 << n);
std::iota(a.begin(), a.end(), 0);
while (a.size() > 1) {
std::vector<int> na;
if (a.size() == 2) {
int r = Q(a[0], a[1]);
assert(r != 0);
if (r == 1)
na.push_back(a[0]);
else
na.push_back(a[1]);
} else {
for (int i = 0; i < a.size(); i += 4) {
int r0 = Q(a[i], a[i + 2]);
if (r0 == 0) {
int r1 = Q(a[i + 1], a[i + 3]);
assert(r1 != 0);
if (r1 == 1) {
na.push_back(a[i + 1]);
} else if (r1 == 2) {
na.push_back(a[i + 3]);
}
} else if (r0 == 1) {
int r1 = Q(a[i], a[i + 3]);
assert(r1 != 0);
if (r1 == 1) {
na.push_back(a[i]);
} else if (r1 == 2) {
na.push_back(a[i + 3]);
}
} else if (r0 == 2) {
int r1 = Q(a[i + 1], a[i + 2]);
assert(r1 != 0);
if (r1 == 1) {
na.push_back(a[i + 1]);
} else if (r1 == 2) {
na.push_back(a[i + 2]);
}
}
}
}
a = na;
}
int ans = a[0];
std::cout << "! " << ans + 1 << std::endl;
#ifdef BACKLIGHT
logd(std, ans);
assert(std == ans);
#endif
}
E. Cross Swapping
比赛的时候剩半个小时做这题，没搞出来，甚至思路都完全不对。
对于\(a_{i, j}\)，它要么保持不动，要么只能换到\(a_{j, i}\)。然后就可以将矩阵划分成左上角和右下角两个部分，后续只需要考虑左上角。
\(a_{i, j}\)保持不动需要\(k = i\)和\(k = j\)这两个操作同时操作，或者同时不操作。换到\(a_{j, i}\)需要只操作\(k = i\)和\(k = j\)这两个操作其中一个。对于\(a_{i, j}\)，看它和\(a_{j, i}\)的关系，可以判断出\(a_{i, j}\)换到\(a_{j, i}\)上会不会更优。
把每个\(i\)分为两个状态：操作和不操作。如果\(a_{i, j} < a_{j, i}\)的话，就说明\(i\)和\(j\)状态不能一样；如果\(a_{i, j} > a_{j, i}\)的话，就说明\(i\)和\(j\)状态需要一样。根据这个关系就可以建出图把问题转化成图论问题。现在这个问题有点像2-SAT，只是2-SAT要求不能有冲突，而这个问题出现冲突的时候可以以权重大的优先。
对于建出来的图中的每个连通块，对于其中权重最大的节点\(i\)，它是否操作其实都是可行的。因为对于和节点\(i\)对应的操作，以\(a_{i, j}\)换到\(a_{j, i}\)为例，它操作了那么\(j\)不操作就可以，它不操作那么\(j\)操作就可以。这样子，默认连通块中权重最大的节点不操作的话，节点是否操作就可以用带权并查集维护。
AC代码
// Problem: E. Cross Swapping
// Contest: Codeforces - Codeforces Round #812 (Div. 2)
// URL: https://codeforces.com/contest/1713/problem/E
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define CPPIO std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
#define freep(p) p ? delete p, p = nullptr, void(1) : void(0)
#ifdef BACKLIGHT
#include "debug.h"
#else
#define logd(...) ;
#endif
using i64 = int64_t;
using u64 = uint64_t;
void solve_case(int Case);
int main(int argc, char* argv[]) {
CPPIO;
int T = 1;
std::cin >> T;
for (int t = 1; t <= T; ++t) {
solve_case(t);
}
return 0;
}
class WDSU {
public:
WDSU(int n) : n_(n), f_(n_), type_(n_) {
for (int i = 0; i < n_; ++i) {
f_[i] = i;
type_[i] = 0;
}
}
int find(int x) {
if (x != f_[x]) {
int fx = f_[x];
int ffx = find(fx);
f_[x] = ffx;
type_[x] = type_[x] ^ type_[fx];
}
return f_[x];
}
void merge(int x, int y, int z) {
int fx = find(x), fy = find(y);
if (fx == fy)
return;
f_[fy] = fx;
type_[fy] = type_[x] ^ type_[y] ^ z;
}
int type(int x) {
int fx = find(x);
return type_[x];
}
private:
int n_;
std::vector<int> f_;
std::vector<int> type_;
};
void solve_case(int Case) {
int n;
std::cin >> n;
std::vector<std::vector<int>> a(n, std::vector<int>(n));
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
std::cin >> a[i][j];
}
}
WDSU d(n);
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
if (a[i][j] < a[j][i]) {
d.merge(i, j, 0);
}
if (a[i][j] > a[j][i]) {
d.merge(i, j, 1);
}
}
}
for (int i = 0; i < n; ++i) {
if (d.type(i) == 1) {
for (int j = 0; j < n; ++j)
std::swap(a[i][j], a[j][i]);
}
}
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
std::cout << a[i][j] << " \n"[j + 1 == n];
}
}
}
F. Lost Array
To be solved.
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