题会慢慢补完的，最近在帮别人上课，补题速度很慢，一边看ppt一边刷ppt的题还要补多校的，感觉有点应付不过来，不过以后会慢慢补完的（指整个暑假）

A

这场出题人的英文水平真的搞事，不说了（（（
这题起码读了X遍才搞明白题意，每次读都是不同的意思，出题人好好表达不行吗（（（
题意是说这个人在长度为N的环上走，等概率随机往左往右，如果全部走完一次就离开，问在M点离开的概率。
再次吐槽，这个N也真的搞事，有时用来表达N steps有时又是长度为N，而且输出还是前缀和
考虑吸收壁的随机游走，然后我们改一下，改成环，吸收条件也改一下，但是这依然可以感受出来极限情况下除了0不能作为结束之外等概率（严格证起来好像很日，然后就完了

/* ***********************************************
Author        :BPM136
Created Time  :7/21/2019 7:24:02 PM
File Name     :A.cpp
************************************************ */#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<iomanip>
#include<bitset>
#include<queue>
#include<ctime>
#include<set>
#include<map>
#include<list>
#include<vector>
#include<cassert>
#include<string>
#include<cinttypes>
#include<cstdint>
#include<array>
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define USE_CIN_COUT ios::sync_with_stdio(0)using namespace std;typedef long long ll;
typedef double db;
typedef long double ld;
typedef unsigned int ui;
typedef unsigned long long ull;ll const mod = 1e9 + 7;ll KSM(ll a, ll k) {ll ret = 1;for (; k; k >>= 1, a = a * a % mod)if (k & 1)ret = ret * a % mod;return ret;
}int main() {USE_CIN_COUT;int T, n, m;cin >> T;ll ans = 1;for (int i = 1; i <= T; ++i) {cin >> n >> m;if (n != 1) {if (m == 0)ans = 0;elseans = ans * KSM(n - 1, mod - 2) % mod;}cout << ans << '\n';}return 0;
}

B

C

D

显然大团是包含小团的，不过比赛的时候上头了没考虑怎么优化，只想着卡常（（（
其实只要把bfs的时候让1的赋予顺序是从左往右就行，这样就从 O ( K n l o g ( K n ) ) O(Knlog(Kn)) O(Knlog(Kn))变成了 O ( K l o g ( K n ) ) O(Klog(Kn)) O(Klog(Kn))
以及判断的时候可以压位，这样就 O ( N 64 ) O(\frac{N}{64}) O(64N)的判断时间了（就是这个给了我盲目的信心想大力爆过去（（（（

/* ***********************************************
Author        :BPM136
Created Time  :7/20/2019 1:35:41 PM
File Name     :D.cpp
************************************************ */#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<iomanip>
#include<bitset>
#include<queue>
#include<ctime>
#include<set>
#include<map>
#include<list>
#include<vector>
#include<cassert>
#include<string>
#include<cinttypes>
#include<cstdint>
#include<array>
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define USE_CIN_COUT ios::sync_with_stdio(0)using namespace std;typedef long long ll;
typedef double db;
typedef long double ld;
typedef unsigned int ui;
typedef unsigned long long ull;int const N = 105;string e[N];
ll e_st1[N], e_st2[N];
int a[N];
int n, K;struct node {ll st1, st2;ll val;node() {}node(ll _st1, ll _st2, ll _val) : st1(_st1), st2(_st2), val(_val) {}bool operator<(node const& oth) const {return val > oth.val;}
};inline void getSt(ll st1, ll st2, ll& nst1, ll& nst2, int np) {nst1 = st1;nst2 = st2;if (np < 50)nst1 |= 1LL << np;elsenst2 |= 1LL << (np - 50);
}inline node getNewNode(node const& x, int np) {node ret = x;if (np < 50)ret.st1 |= 1LL << np;elseret.st2 |= 1LL << (np - 50);ret.val += a[np];return ret;
}inline bool pInSta(ll const& st1, ll const& st2, int p) {if (p < 50)return st1 & (1LL << p);elsereturn st2 & (1LL << (p - 50));
}inline bool eInSta(ll const& st1, ll const& st2, ll const& e1, ll const& e2) {return ((st1 & e1) == st1) && ((st2 & e2) == st2);
}//set<node> S;int main() {double tt = clock();freopen("in.txt", "r", stdin);cin >> n >> K;--K;for (int i = 0; i < n; ++i)cin >> a[i];for (int i = 0; i < n; ++i)cin >> e[i];for (int i = 0; i < n; ++i)for (int j = 0; j < n; ++j)if (e[i][j] == '1')getSt(e_st1[i], e_st2[i], e_st1[i], e_st2[i], j);//bitset<5> debug1, debug2;auto Q = priority_queue<node>();Q.push(node{ 0, 0, 0 });// S.insert(node{ 0, 0, 0 });node tmp;while (Q.empty() == 0 && K--) {auto now = Q.top();Q.pop();//debug1 = now.st1, debug2 = now.st2;// cerr << debug1 << ' ' << debug2 << ' ' << now.val << ' ' << Q.size() << ' ' << K << '\n';int mx = 0;for (int i = 0; i < n; ++i)if (pInSta(now.st1, now.st2, i))mx = i + 1;for (int i = mx; i < n; ++i)if (eInSta(now.st1, now.st2, e_st1[i], e_st2[i])) {tmp = getNewNode(now, i);//if (S.count(tmp) == 0) {Q.push(tmp);//    S.insert(tmp);//}}}// cerr << K << ' ' << Q.empty() << ' ' << Q.top().val << '\n';if (K >= 0 || Q.empty())cout << -1 << '\n';elsecout << Q.top().val << '\n';cerr << clock() - tt << '\n';return 0;
}

E

F

dfs

#include <iostream>
#include <limits>
#include <cstdint>
#include <algorithm>
#include <string>
#include <vector>using i64 = int64_t;
const int N = 14 * 2;int n, v[N][N];
int ss1[N], ss2[N];
i64 ans, cur;void dfs (int i, int cnt1) {if (ans <= cur)return;if (i == n) {ans = cur;return;}if (cnt1 < (n >> 1)) {i64 s1 = 0;for (int j = 0; j < cnt1; ++j)s1 += v[i][ss1[j]];ss1[cnt1] = i;// std::cerr << "> " << i << " " << s1 << "\n";cur += s1;dfs(i + 1, cnt1 + 1);cur -= s1;}int cnt2 = i - cnt1;if (cnt2 < (n >> 1)) {i64 s2 = 0;for (int j = 0; j < cnt2; ++j)s2 += v[i][ss2[j]];ss2[cnt2] = i;cur += s2;dfs(i + 1, cnt1);cur -= s2;}
}int main () {std::ios::sync_with_stdio(false);std::cin >> n;n *= 2;i64 sum = 0;for (int i = 0; i < n; ++i)for(int j = 0; j < n; ++j) {std::cin >> v[i][j];if (i < j)sum += v[i][j];}ans = std::numeric_limits<i64>::max();cur = 0;dfs(0, 0);// std::cerr << "s = " << ans << '\n';std::cout << sum - ans << '\n';return 0;
}

G

H

原题，单调栈爆过去就行

/* ***********************************************
Author        :BPM136
Created Time  :7/20/2019 12:41:43 PM
File Name     :H.cpp
************************************************ */#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<iomanip>
#include<bitset>
#include<queue>
#include<ctime>
#include<set>
#include<map>
#include<list>
#include<vector>
#include<cassert>
#include<string>
#include<cinttypes>
#include<cstdint>
#include<array>
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define USE_CIN_COUT ios::sync_with_stdio(0)using namespace std;typedef long long ll;
typedef double db;
typedef long double ld;
typedef unsigned int ui;
typedef unsigned long long ull;int const N = 1005;int lef[N], sta[N];
int row[N][N];
int val[N][N];
int a[N][N];
int n, m;void update(int& ans1, int& ans2, int tmp) {if (tmp >= ans1) {ans2 = ans1;ans1 = tmp;} elseif (tmp > ans2) ans2 = tmp;
}int main() {cin >> n >> m;string s;for (int i = 1; i <= n; ++i) {cin >> s;for (int j = 1; j <= m; ++j) {a[i][j] = s[j - 1] == '1';if (a[i][j]) a[i][j] = a[i][j - 1] + 1;}}auto ans1 = 0, ans2 = 0;for (int j = 1; j <= m; ++j) {int top = 0;a[n + 1][j] = -1;for (int i = 1; i <= n + 1; ++i) {if (top == 0 || a[i][j] > a[sta[top - 1]][j]) {sta[top++] = i;lef[i] = i;continue;}if (a[i][j] == a[sta[top - 1]][j])continue;while (top >= 1 && a[i][j] < a[sta[top - 1]][j]) {--top;update(ans1, ans2, (i - lef[sta[top]]) * a[sta[top]][j]);update(ans1, ans2, (i - lef[sta[top]] - 1) * a[sta[top]][j]);update(ans1, ans2, (i - lef[sta[top]]) * (a[sta[top]][j] - 1));}lef[i] = lef[sta[top]];sta[top++] = i;}}cout << ans2 << endl;
}

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2019 牛客多校第二场 4/10

A

B

C

D

E

F

G

H

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