标签：inf 目录 std pac include 攻击 amp += int
| Solved | A | B | C | D | E | F | G | H | I |
|---|---|---|---|---|---|---|---|---|---|
| 5 / 9 | O | - | ? | ? | - | O | - | - | O |
简单博弈，显然每次每个人只会选择一个。
/*
* Author: heyuhhh
* Created Time: 2020/5/23 13:16:27
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
#define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
void err() { std::cout << std::endl; }
template<typename T, typename...Args>
void err(T a, Args...args) { std::cout << a << ‘ ‘; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) {
for (auto &v : arg) std::cout << v << ‘ ‘; err(args...); }
#else
#define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 1e5 + 5;
void run() {
int n, m;
cin >> n;
for (int i = 1; i <= n; i++) {
string s; cin >> s;
}
cin >> m;
for (int i = 1; i <= m; i++) {
string s; cin >> s;
}
if (n <= m) cout << "Quber CC" << ‘\n‘;
else cout << "Cuber QQ" << ‘\n‘;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
run();
return 0;
}
题意：
随机在三维空间中给定\(n\)个点，现在有病毒进行攻击，每次病毒会攻击当前没有受保护的点。
定义结点\(i\)保护结点\(j\)为：
问病毒攻击几轮能消灭所有的点，并且输出每个点存活的轮数。
思路：
因为是随机数据，所以我们可以直接通过模拟上述过程来做。
模拟的思想就是我们对于每一轮攻击，选出一个最小的点，然后找到所有不能保护的点，他们都会在这一轮攻击中被消灭；然后进行下一轮攻击。
具体实现我们可以直接枚举或者用树状数组维护\(z\)的最小值都可。反正数据随机可以不用管复杂度= =
还有一种复杂度比较稳定的做法就是cdq分治，用cdq分治维护每个结点的拓扑序即可。注意一下这种写法我们要按照拓扑序进行转移，也就是说在cdq分治中，先处理左边的，然后处理左边对右边的影响，最后再处理右边。这样能够保证拓扑序从小到大转移。
详见代码：
/*
* Author: heyuhhh
* Created Time: 2020/5/23 14:57:55
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
#define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
void err() { std::cout << std::endl; }
template<typename T, typename...Args>
void err(T a, Args...args) { std::cout << a << ‘ ‘; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) {
for (auto &v : arg) std::cout << v << ‘ ‘; err(args...); }
#else
#define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 1e5 + 5;
unsigned long long k1, k2;
unsigned long long CoronavirusBeats() {
unsigned long long k3 = k1, k4 = k2;
k1 = k4;
k3 ^= k3 << 23;
k2 = k3 ^ k4 ^ (k3 >> 17) ^ (k4 >> 26);
return k2 + k4;
}
unsigned long long x[N], y[N], z[N];
struct Point {
unsigned long long x, y, z;
int id;
}a[N], b[N];
int n;
int ans[N];
void Hash(unsigned long long* a) {
sort(a + 1, a + n + 1);
a[0] = unique(a + 1, a + n + 1) - a - 1;
}
int find(unsigned long long* a, unsigned long long x) {
return lower_bound(a + 1, a + a[0] + 1, x) - a;
}
int c[N];
int lowbit(int x) {
return x & (-x);
}
void del(int x) {
for (int i = x; i < N; i += lowbit(i)) c[i] = 0;
}
void add(int x, int v) {
for(int i = x; i < N; i += lowbit(i)) c[i] = max(c[i], v);
}
int query(int x) {
int ans = 0;
for(int i = x; i; i -= lowbit(i)) ans = max(ans, c[i]);
return ans;
}
void cdq(int l, int r) {
if(l == r) return ;
int mid = (l + r) >> 1;
cdq(l, mid);
sort(a + l, a + mid + 1, [&](Point A, Point B) {
return A.y < B.y;
});
sort(a + mid + 1, a + r + 1, [&](Point A, Point B) {
return A.y < B.y;
});
int t1 = l, t2 = mid + 1;
for(int i = l; i <= r; i++) {
if((t1 <= mid && a[t1].y <= a[t2].y) || t2 > r) {
add(a[t1].z, ans[a[t1].id]);
b[i] = a[t1++];
} else {
ans[a[t2].id] = max(ans[a[t2].id], query(a[t2].z) + 1);
b[i] = a[t2++];
}
}
for(int i = l; i <= mid; i++) del(a[i].z);
sort(a + mid + 1, a + r + 1, [&](Point A, Point B) {
return A.x < B.x;
});
cdq(mid + 1, r);
}
void run() {
cin >> n >> k1 >> k2;
for (int i = 1; i <= n; ++i) {
x[i] = CoronavirusBeats();
y[i] = CoronavirusBeats();
z[i] = CoronavirusBeats();
a[i] = Point{x[i], y[i], z[i], i};
}
Hash(x), Hash(y), Hash(z);
for (int i = 1; i <= n; i++) {
a[i].x = find(x, a[i].x);
a[i].y = find(y, a[i].y);
a[i].z = find(z, a[i].z);
}
sort(a + 1, a + n + 1, [&](Point A, Point B) {
return A.x < B.x;
});
for (int i = 1; i <= n; i++) {
ans[i] = 1;
}
cdq(1, n);
int res = 0;
for (int i = 1; i <= n; i++) {
res = max(res, ans[i]);
--ans[i];
}
cout << res << ‘\n‘;
for (int i = 1; i <= n; i++) {
cout << ans[i] << " \n"[i == n];
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
run();
return 0;
}
题意：
给定一颗带点权的树，定义一条路径的价值为\(\displaystyle \frac{a_{p_1}\cdot a_{p_2}\cdots a_{p_m}}{m}\)。
现在问路径价值的最小值为多少，输出\("x/y"\)。
思路：
分几种清空考虑：
思路就是上面这样，具体的证明应该也比较好证，就拿第三种情况证明一下：
首先答案为\(\frac{1}{m}\)，假设现在有\(t\)段连续的\(1\)长度为\(m\)，那么答案为\(\displaystyle\frac{b_1\cdots b_{t-1}}{tm+t-1}\)，因为\(b_i\geq 2\)，我们就取\(b_i=2\)，将其与\(\frac{1}{m}\)做差得\(\displaystyle m(2^{t-1}-t)-t+1\)。容易发现当\(t=2\)时，差为负数；当\(t>2\)时，差为非负数；若\(b_i\geq 3\)，值为非负数。也就是说只有我们上述所说情况会出现最优解。
证明大概就是这个样子，可能细节会有点问题。。
思路是这样，代码写起来还是挺麻烦的，需要换下根。
详见代码：
/*
* Author: heyuhhh
* Created Time: 2020/5/23 20:42:25
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <bits/stdc++.h>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
#define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
void err() { std::cout << std::endl; }
template<typename T, typename...Args>
void err(T a, Args...args) { std::cout << a << ‘ ‘; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) {
for (auto &v : arg) std::cout << v << ‘ ‘; err(args...); }
#else
#define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 1e6 + 5;
int n;
int a[N];
struct Edge {
int u, v;
}e[N];
vector <int> G[N];
bool vis[N];
int f[N][2], g[N];
void run() {
cin >> n;
for (int i = 1; i < n; i++) {
int u, v; cin >> u >> v;
e[i] = Edge {u, v};
}
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
pii ans(INF, INF);
for (int i = 1; i <= n; i++) {
if (a[i] == 0) {
cout << "0/1" << ‘\n‘;
return;
}
ans = min(ans, MP(a[i], 1));
}
if (ans.fi > 1) {
cout << ans.fi << ‘/‘ << ans.se << ‘\n‘;
return;
}
for (int i = 1; i < n; i++) {
int u = e[i].u, v = e[i].v;
if (a[u] == 1 && a[v] == 1) {
G[u].push_back(v);
G[v].push_back(u);
}
}
int m = 0, rt;
function <void(int, int, int)> dfs;
dfs = [&](int u, int fa, int d) -> void {
if (d > m) {
m = d, rt = u;
}
vis[u] = 1;
for (auto v : G[u]) if (v != fa) {
dfs(v, u, d + 1);
}
};
function <void(int, int)> dfs2;
dfs2 = [&](int u, int fa) -> void {
f[u][0] = 1, f[u][1] = -1;
for (auto v : G[u]) if (v != fa) {
dfs2(v, u);
if (f[v][0] + 1 > f[u][0]) {
f[u][1] = f[u][0];
f[u][0] = f[v][0] + 1;
} else if(f[v][0] + 1 > f[u][1]) {
f[u][1] = f[v][0] + 1;
}
}
};
function <void(int, int, int)> dfs3;
dfs3 = [&](int u, int fa, int h) -> void {
for (auto v : G[u]) if (v != fa) {
g[v] = f[v][0];
if (f[v][0] + 1 == f[u][0]) {
g[v] = max(g[v], max(h, f[u][1]) + 1);
dfs3(v, u, max(h, f[u][1]) + 1);
} else {
g[v] = max(g[v], max(h, f[u][0]) + 1);
dfs3(v, u, max(h, f[u][0]) + 1);
}
}
};
int MAX = 0;
for (int i = 1; i <= n; i++) if (a[i] == 1 && !vis[i]) {
rt = i, m = 0;
dfs(rt, 0, 1), dfs(rt, 0, 1);
MAX = max(m, MAX);
dfs2(i, 0);
g[i] = f[i][0];
dfs3(i, 0, 0);
}
ans = MP(1, MAX);
vector <int> d(n + 1);
for (int i = 1; i < n; i++) {
int u = e[i].u, v = e[i].v;
if (a[u] > a[v]) swap(u, v);
if (a[u] == 1 && a[v] == 2) {
if (g[u] == MAX) ++d[v];
if (d[v] >= 2) {
ans = MP(2, 2 * MAX + 1);
break;
}
}
}
cout << ans.fi << ‘/‘ << ans.se << ‘\n‘;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
run();
return 0;
}
排序过后判断是否为某一个的前缀就行，我的时间复杂度貌似有点炸，但依旧跑得飞快。
正解的话就直接上字典树吧。
/*
* Author: heyuhhh
* Created Time: 2020/5/23 13:35:27
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
#define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
void err() { std::cout << std::endl; }
template<typename T, typename...Args>
void err(T a, Args...args) { std::cout << a << ‘ ‘; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) {
for (auto &v : arg) std::cout << v << ‘ ‘; err(args...); }
#else
#define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 1e6 + 5;
void run() {
int n; cin >> n;
vector <string> s(n);
for (int i = 0; i < n; i++) {
cin >> s[i];
}
sort(all(s));
int ans = 0;
auto chk = [&] (string& str) {
int len = str.length();
if (len > 3 && str.substr(len - 4) == ".eoj") return 1;
return 0;
};
for (int i = 0, j; i < n; i++) {
if (i == n - 1) {
ans += chk(s[i]);
} else {
int l1 = s[i].length(), l2 = s[i + 1].length();
if (l1 > l2) {
ans += chk(s[i]);
} else if (s[i + 1].substr(0, l1) != s[i]) {
ans += chk(s[i]);
}
}
}
cout << ans << ‘\n‘;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
run();
return 0;
}
随便构造一下就行。
/*
* Author: heyuhhh
* Created Time: 2020/5/23 13:23:08
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
#define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
void err() { std::cout << std::endl; }
template<typename T, typename...Args>
void err(T a, Args...args) { std::cout << a << ‘ ‘; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) {
for (auto &v : arg) std::cout << v << ‘ ‘; err(args...); }
#else
#define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 1e5 + 5;
void run() {
int n, k; cin >> n >> k;
string ans = "";
for (int i = 1; i < k; i++) ans += "a";
for (int i = 1; i <= n - k; i++) ans += "c";
ans += "b";
reverse(all(ans));
cout << ans << ‘\n‘;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
run();
return 0;
}
标签：inf 目录 std pac include 攻击 amp += int
原文地址：https://www.cnblogs.com/heyuhhh/p/12945175.html
