视频题解。
注意到\(n\leq 1000\)，所以直接暴力枚举即可。
/*
* Author: heyuhhh
* Created Time: 2020/5/31 22:35:54
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#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, x; cin >> n >> x;
vector <int> a(n);
int cnt[2] = {0, 0};
for (int i = 0; i < n; i++) {
cin >> a[i];
if (a[i] & 1) ++cnt[1];
else ++cnt[0];
}
for (int i = 1; i <= x; i += 2) {
if (cnt[1] >= i && cnt[0] >= x - i) {
cout << "YES" << ‘\n‘;
return;
}
}
cout << "NO" << ‘\n‘;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
int T; cin >> T; while(T--)
run();
return 0;
}
最后的形式一定为\(000...111,111...000,1111..11,00..00\)这几种形式，所以只用考虑这几种情况就行。前两种枚举一下即可解决，后两种也很简单。
/*
* Author: heyuhhh
* Created Time: 2020/5/31 22:42:37
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#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() {
string s; cin >> s;
int n = s.length();
int ans = INF;
vector <int> sum(n);
for (int i = 0; i < n; i++) {
if (i == 0) sum[0] = (s[i] == ‘1‘);
else sum[i] = sum[i - 1] + (s[i] == ‘1‘);
}
int cnt0 = 0, cnt1 = 0;
for (int i = n - 1; i; i--) {
if (s[i] == ‘0‘) {
++cnt0;
} else {
++cnt1;
}
ans = min(ans, min(sum[i - 1], i - sum[i - 1]) + min(cnt0, cnt1));
}
ans = min(ans, min(sum[n - 1], n - sum[n - 1]));
cout << ans << ‘\n‘;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
int T; cin >> T; while(T--)
run();
return 0;
}
首先\(x\)为叶子结点(注意\(n=1\)的情况)必胜。
否则最后必胜态为如下形式：
具体证明可参见题解或者视频里所说的。
容易发现最终答案只与结点个数的奇偶性有关，所以判断一下就行。
/*
* Author: heyuhhh
* Created Time: 2020/5/31 23:06:05
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#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, x; cin >> n >> x; --x;
vector <int> d(n);
for (int i = 1; i < n; i++) {
int u, v; cin >> u >> v;
--u, --v;
++d[u], ++d[v];
}
if (d[x] <= 1) {
cout << "Ayush" << ‘\n‘;
return;
}
if (n & 1) cout << "Ashish" << ‘\n‘;
else cout << "Ayush" << ‘\n‘;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
int T; cin >> T; while(T--)
run();
return 0;
}
交互题。
存在一种刚好用\(12\)次的方法，具体方法可以看看视频题解，注意第一次询问我们直接对两个集合进行询问，同样能找到最大值，并且还可以在分治的时候省略一层。
细节见代码：
/*
* Author: heyuhhh
* Created Time: 2020/5/31 23:56:07
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#include <random>
#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;
mt19937 rnd(time(NULL));
int n, k;
int query2(vector <int>& a) {
cout << "? " << sz(a) << ‘\n‘;
for (auto it : a) cout << it << ‘ ‘;
cout << endl;
int x; cin >> x;
return x;
}
int query(vector <int> a) {
sort(all(a));
vector <int> b;
int j = 0;
for (int i = 1; i <= n; i++) {
if (j >= sz(a) || i != a[j]) {
b.push_back(i);
} else ++j;
}
return query2(b);
}
void run() {
cin >> n >> k;
vector <int> ans(k);
vector <vector<int>> s(k);
for (int i = 0; i < k; i++) {
int c; cin >> c;
s[i].resize(c);
for (int j = 0; j < c; j++) {
cin >> s[i][j];
}
sort(all(s[i]));
}
int Max;
int m = (k - 1) / 2;
vector <int> a, b;
for (int i = 0; i <= m; i++) {
for (auto it : s[i]) {
a.push_back(it);
}
}
for (int i = m + 1; i < k; i++) {
for (auto it : s[i]) {
b.push_back(it);
}
}
int Max1 = query(a), Max2 = query(b);
Max = max(Max1, Max2);
if (Max1 == Max2) {
for (int i = 0; i < k; i++) {
ans[i] = Max1;
}
} else {
function <int(int, int)> solve;
solve = [&] (int l, int r) -> int {
if (l == r) return l;
int m = (l + r) >> 1;
vector <int> a;
for (int i = l; i <= m; i++) {
for (auto it : s[i]) {
a.push_back(it);
}
}
int now = query(a);
if (now == Max) return solve(m + 1, r);
return solve(l, m);
};
int t;
if (Max1 < Max2) t = solve(0, m);
else t = solve(m + 1, k - 1);
for (int i = 0; i < k; i++) {
if (i != t) ans[i] = Max;
else ans[i] = query(s[i]);
}
}
cout << "!";
for (auto it : ans) cout << ‘ ‘ << it;
cout << endl;
string ok; cin >> ok;
if (ok != "Correct") exit(0);
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
int T; cin >> T; while(T--)
run();
return 0;
}
注意到\(a[u]=min(a[u],a[fa[u]])\)，那么执行完这个操作过后问题将变为一个子树内的问题，与父亲结点无关。
在一个子数内的话直接贪心将儿子结点剩余进行匹配即可。
/*
* Author: heyuhhh
* Created Time: 2020/5/31 23:20:34
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#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 = 2e5 + 5;
int n;
vector <int> G[N];
int a[N], b[N], c[N];
int cnt[N][2];
ll ans;
void dfs(int u, int fa) {
if (fa) a[u] = min(a[u], a[fa]);
int f = -1;
if (b[u] == 0 && c[u]) f = 0;
if (b[u] && c[u] == 0) f = 1;
int c0 = 0, c1 = 0;
for (auto v : G[u]) if (v != fa) {
dfs(v, u);
cnt[u][0] += cnt[v][0];
cnt[u][1] += cnt[v][1];
if (cnt[v][0] > cnt[v][1]) c0 += cnt[v][0] - cnt[v][1];
else c1 += cnt[v][1] - cnt[v][0];
}
if (f == 0) ++cnt[u][0], ++c0;
if (f == 1) ++cnt[u][1], ++c1;
ans += 2ll * a[u] * min(c0, c1);
}
void run() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i] >> b[i] >> c[i];
}
for (int i = 1; i < n; i++) {
int u, v; cin >> u >> v;
G[u].push_back(v);
G[v].push_back(u);
}
dfs(1, 0);
if (cnt[1][0] != cnt[1][1]) ans = -1;
cout << ans << ‘\n‘;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
run();
return 0;
}
很神仙的dp，不好想。
有两种做法，第一种做法可以参见视频题解里面说的，我们要求一个最长的公共子序列，并且剩下的每一个在\(s\)串中没有匹配的数都只能往前移动。
那么即是对每一个匹配的\(s_i,t_j\),\(i+1\rightarrow n\)中\(x\)字符出现的次数不小于\(t+1\rightarrow n\)中出现的次数，这样的话最终剩下没有匹配的只能往右边走。那么最终答案即为\(n-dp[n][n]\)。
另外一种做法也很神奇！我们定义\(dp[i][j]\)为\(s\)串中确定了\(i\)个字符，\(t\)串匹配到了\(j\)时，最少需要插入的数的个数，显然最终答案即为\(dp[n][n]\)。
那么我们将题目中的操作看作两部分，一部分是选出一些数准备插入，另外一部分就是插入。
那么往后移动\(j\)时其实我们在不断选择一些数出来，然后往后移动\(i\)时我们在不断插入一些数，代价为\(1\)，当然如果\(s[i]=t[j]\)的话就不用消耗代价。
细节见代码吧。。
/*
* Author: heyuhhh
* Created Time: 2020/6/1 13:32:17
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#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 = 2000 + 5;
void run() {
int n; cin >> n;
string s, t; cin >> s >> t;
string ss = s, tt = t;
sort(all(ss)), sort(all(tt));
if (ss != tt) {
cout << -1 << ‘\n‘;
return;
}
s = ‘#‘ + s;
t = ‘#‘ + t;
vector <vector <int>> dp(n + 1, vector<int>(n + 1));
vector <vector <int>> suf_s(n + 2, vector<int>(26)), suf_t(n + 2, vector<int>(26));
for (int i = n; i >= 1; i--) {
for (int j = 0; j < 26; j++) {
suf_s[i][j] = suf_s[i + 1][j];
suf_t[i][j] = suf_t[i + 1][j];
}
++suf_s[i][s[i] - ‘a‘];
++suf_t[i][t[i] - ‘a‘];
}
dp[0][0] = 0;
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= n; j++) {
if (i) dp[i][j] = dp[i - 1][j];
if (j) dp[i][j] = max(dp[i][j], dp[i][j - 1]);
if (i && j && s[i] == t[j]) {
bool ok = true;
for (int k = 0; k < 26; k++) {
if (suf_s[i][k] < suf_t[j][k]) ok = false;
}
if (ok) {
dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);
}
}
}
}
cout << n - dp[n][n] << ‘\n‘;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
int T; cin >> T; while(T--)
run();
return 0;
}
/*
* Author: heyuhhh
* Created Time: 2020/6/1 21:01:21
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#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; cin >> n;
string s, t; cin >> s >> t;
string ss = s, tt = t;
sort(all(ss)), sort(all(tt));
if (ss != tt) {
cout << -1 << ‘\n‘;
return;
}
s = ‘#‘ + s;
t = ‘#‘ + t;
vector <vector <int>> dp(n + 1, vector<int>(n + 1, INF));
vector <vector <int>> suf_s(n + 2, vector<int>(26)), suf_t(n + 2, vector<int>(26));
for (int i = n; i >= 1; i--) {
for (int j = 0; j < 26; j++) {
suf_s[i][j] = suf_s[i + 1][j];
suf_t[i][j] = suf_t[i + 1][j];
}
++suf_s[i][s[i] - ‘a‘];
++suf_t[i][t[i] - ‘a‘];
}
for (int i = 0; i <= n; i++) {
dp[0][i] = 0;
}
for (int i = 1; i <= n; i++) {
for (int j = i; j <= n; j++) {
if (s[i] == t[j]) {
dp[i][j] = dp[i - 1][j - 1];
}
dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1);
if (suf_s[i + 1][t[j] - ‘a‘] > suf_t[j + 1][t[j] - ‘a‘]) {
dp[i][j] = min(dp[i][j], dp[i][j - 1]);
}
}
}
cout << dp[n][n] << ‘\n‘;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
int T; cin >> T; while(T--)
run();
return 0;
}
Codeforces Round #646 (Div. 2)
原文：https://www.cnblogs.com/heyuhhh/p/13027492.html