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EntropyIncreaserEntropyIncreaser
committed
2019.11.23
1 parent 30564f4 commit 5f0d2fc

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3 files changed

+213
-2
lines changed

3 files changed

+213
-2
lines changed

‎part2/agc030_f.cpp

+117
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@@ -0,0 +1,117 @@
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#include <cstdio>
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#include <cstring>
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#include <cstdlib>
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#include <cmath>
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#include <ctime>
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#include <cctype>
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#include <algorithm>
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#include <random>
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#include <bitset>
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#include <queue>
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#include <functional>
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#include <set>
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#include <map>
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#include <vector>
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#include <chrono>
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#include <iostream>
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#include <limits>
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#include <numeric>
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#define LOG(FMT...) fprintf(stderr, FMT)
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using namespace std;
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typedef long long ll;
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typedef unsigned long long ull;
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// mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
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template <class T>
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istream& operator>>(istream& is, vector<T>& v) {
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for (T& x : v)
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is >> x;
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return is;
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}
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template <class T>
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ostream& operator<<(ostream& os, const vector<T>& v) {
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if (!v.empty()) {
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os << v.front();
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for (int i = 1; i < v.size(); ++i)
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os << ' ' << v[i];
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}
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return os;
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}
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const int P = 1000000007;
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void add(int& x, int y) {
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if ((x += y) >= P)
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x -= P;
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}
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void sub(int& x, int y) {
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if ((x -= y) < 0)
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x += P;
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}
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int norm(int x) { return x >= P ? x - P : x; }
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int main() {
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#ifdef LBT
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freopen("test.in", "r", stdin);
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int nol_cl = clock();
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#endif
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ios::sync_with_stdio(false);
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cin.tie(nullptr);
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int n;
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cin >> n;
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vector<int> a(n * 2), pos(n * 2, -1);
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cin >> a;
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for (int i = 0; i < n * 2; ++i)
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if (a[i] != -1)
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pos[--a[i]] = i;
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int empties = 0;
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for (int i = 0; i < n; ++i)
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if (a[i * 2] == -1 && a[i * 2 + 1] == -1)
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++empties;
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vector<vector<int>> dp(n + 1, vector<int>(n + 1));
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dp[0][0] = 1;
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for (int i = 0; i < n * 2; ++i) {
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if (pos[i] != -1 && a[pos[i] ^ 1] != -1)
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continue;
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vector<vector<int>> newDp(n + 1, vector<int>(n + 1));
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if (pos[i] == -1) {
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for (int j = 0; j <= n; ++j)
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for (int k = 0; k <= n; ++k) {
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if (k < n)
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add(newDp[j][k + 1], dp[j][k]);
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if (j < n)
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add(newDp[j + 1][k], dp[j][k]);
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if (k)
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add(newDp[j][k - 1], dp[j][k]);
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}
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} else {
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for (int j = 0; j <= n; ++j)
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for (int k = 0; k <= n; ++k) {
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if (k < n)
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add(newDp[j][k + 1], dp[j][k]);
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if (j)
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add(newDp[j - 1][k], dp[j][k] * (ll)j % P);
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}
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}
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dp = newDp;
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}
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int ans = dp[0][0];
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for (int i = 1; i <= empties; ++i)
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ans = ans * (ll)i % P;
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cout << ans << '\n';
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#ifdef LBT
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LOG("Time: %dms\n", int ((clock()
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-nol_cl) / (double)CLOCKS_PER_SEC * 1000));
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#endif
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return 0;
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}

‎part2/cf605E.cpp

+92
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#include <cstdio>
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#include <cstring>
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#include <cstdlib>
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#include <cmath>
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#include <ctime>
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#include <cctype>
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#include <algorithm>
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#include <random>
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#include <bitset>
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#include <queue>
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#include <functional>
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#include <set>
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#include <map>
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#include <vector>
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#include <chrono>
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#include <iostream>
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#include <limits>
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#include <numeric>
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#define LOG(FMT...) fprintf(stderr, FMT)
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using namespace std;
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typedef long long ll;
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typedef unsigned long long ull;
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// mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
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template <class T>
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istream& operator>>(istream& is, vector<T>& v) {
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for (T& x : v)
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is >> x;
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return is;
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}
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template <class T>
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ostream& operator<<(ostream& os, const vector<T>& v) {
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if (!v.empty()) {
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os << v.front();
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for (int i = 1; i < v.size(); ++i)
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os << ' ' << v[i];
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}
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return os;
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}
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const int N = 1010;
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int n;
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double p[N][N];
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bool vis[N];
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double ans[N], rest[N], sum[N];
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int main() {
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#ifdef LBT
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freopen("test.in", "r", stdin);
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int nol_cl = clock();
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#endif
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ios::sync_with_stdio(false);
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cin.tie(nullptr);
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scanf("%d", &n);
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for (int i = 1; i <= n; ++i)
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for (int j = 1; j <= n; ++j) {
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int k;
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scanf("%d", &k);
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p[i][j] = k * 0.01;
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}
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fill(rest + 1, rest + n + 1, 1);
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fill(ans + 1, ans + n, 1e9);
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for (int rep = 1; rep < n; ++rep) {
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int u = -1;
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for (int i = 1; i <= n; ++i)
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if (!vis[i] && (u == -1 || ans[u] > ans[i]))
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u = i;
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vis[u] = true;
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for (int i = 1; i <= n; ++i)
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if (!vis[i]) {
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sum[i] += rest[i] * p[i][u] * ans[u];
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rest[i] *= (1 - p[i][u]);
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if (rest[i] != 1)
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ans[i] = (1 + sum[i]) / (1 - rest[i]);
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}
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}
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printf("%.15f\n", ans[1]);
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#ifdef LBT
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LOG("Time: %dms\n", int ((clock()
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-nol_cl) / (double)CLOCKS_PER_SEC * 1000));
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#endif
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return 0;
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}

‎part2/checklist.md

+4-2
Original file line numberDiff line numberDiff line change
@@ -44,7 +44,8 @@
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- [x] agc030_d
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- 记 $f(x, y, 0/1)$ 表示 $(x, y)$ 位置组成的是严格顺序对或逆序对的概率。
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- [ ] agc030_e
47-
- [ ] agc030_f
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- [x] agc030_f
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- 考虑从小到大决定每个数的位置，那么一个数如果作为最小值要么被放在一个没有被钦定过数的格子里，要么被放在一个被钦定了一个数的格子里，考虑在最初分类的时候把空的先视为无序，$f(i, j)$ 表示还剩下 $i$ 个空的格子已经被放了一个数，$j$ 个数确定要被放在和别的钦定的数在一起但是还没确定具体位置。
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- [x] agc031_d
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- 可以发现答案可以表示为一个 $a^{\lfloor k / 6\rfloor} b_{k \bmod 6} a^{-\lfloor k / 6\rfloor}$。
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- [ ] agc031_e
@@ -176,7 +177,8 @@
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- [ ] cf594E
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- [x] cf603E
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- LCT 维护最小生成森林顺便子树大小，另外每次看权值最大的边能不能删掉。
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- [ ] cf605E
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- [x] cf605E
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- 假设每个点的期望答案以及确定，那么策略显然是到开启的虫洞能到达的最早的位置，因此每次维护预计答案，每轮迭代会确定一个新的点，用来更新未确定的点即可，要维护一些中间信息来保证复杂度。
180182
- [ ] cf607E
181183
- [ ] cf611G
182184
- [x] cf611H

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