2022 ICPC Shenyang

2022 ICPC Shenyang

C

区间越大越好、答案区间的左右端点一定可以是已有的数，然后就可以 n 方做了

E

不太懂啊，虽然过了

i64 pw2[N * N + 5];std::vector<int> adj[N + 5];
int d[N + 5], back[N + 5]; // back 表示 i 的子树下有多少点与 i 的祖先有边bool vis[N + 5];
int sz[N + 5];
i64 f[N + 5][N + 5], g[N + 5];void add(i64 &a, i64 b) {a += b;if (a >= Mod) {a -= Mod;}
}void pre(int cur, int lst) {d[cur] = d[lst] + 1;for (int &to : adj[cur]) {if (to == lst) {continue;}if (d[to] == 0) {pre(to, cur);back[cur] += back[to];}else if (d[to] < d[cur]) {back[cur] += 1;back[to] -= 1;}}
}void dfs(int cur, int lst) {vis[cur] = true;sz[cur] = 1;f[cur][1] = 1;for (auto &to : adj[cur]) {if (to == lst || vis[to]) {continue;}dfs(to, cur);for (int i = 1; i <= sz[cur]; ++i) {g[i] = f[cur][i];f[cur][i] = 0;}for (int i = 1; i <= sz[cur]; ++i) {for (int j = 1; j <= sz[to]; ++j) {// i * j - 1 是两个连通块之间可连的边数// -1 是因为 cur - to 已经是连起来的了add(f[cur][i + j], g[i] * f[to][j] % Mod * pw2[i * j - 1] % Mod);}}// 容斥if (back[to] == 0) {i64 sum = 0;for (int i = 1; i <= sz[to]; ++i) {add(sum, f[to][i]);}sum = Mod - sum;for (int i = 1; i <= sz[cur]; ++i) {add(f[cur][i], sum * g[i] % Mod);}}sz[cur] += sz[to];}
}void solve() {int n = 0, m = 0;std::cin >> n >> m;for (int i = 0; i < m; ++i) {int u = 0, v = 0;std::cin >> u >> v;adj[u].push_back(v);adj[v].push_back(u);}pre(1, 0);dfs(1, 0);i64 ans = 0;for (int i = 1; i <= n; ++i) {add(ans, f[1][i]);// debug("%lld ", f[1][i]);}i64 inv2 = (Mod + 1) / 2;// 消除掉非树边的影响for (int i = 0; i < m - (n - 1); ++i) {(ans *= inv2) %= Mod;}std::cout << ans << '\n';
}

L

直接爆搜即可：

std::array<int, 2> n;
// s[0/1][i] 第 0/1 个阵营里面第 i 个人的状态，0 代表血量，1 代表攻击次数
std::array<int, 2> s[2][N];
int atk[2][N];// u 代表当前攻击的人
std::array<long double, 3> dfs(int u) {std::array<long double, 3> res = { 0.0, 0.0, 0.0 };int v = u ^ 1; // 被攻击的人std::array<int, 2> cnt = { 0, 0 };for (int i = 0; i < 2; ++i) {for (int j = 0; j < n[i]; ++j) {if (s[i][j][0] > 0) {cnt[i] += 1;}}}// 平局if (!cnt[0] && !cnt[1]) {res[2] = 1.0;return res;}// 1 胜利if (!cnt[0]) {res[1] = 1;return res;}// 0 胜利if (!cnt[1]) {res[0] = 1;return res;}// 找 u 中攻击次数的最小值int mn = Inf;for (int i = 0; i < n[u]; ++i) {if (s[u][i][0] > 0) {mn = std::min(mn, s[u][i][1]);}} for (int i = 0; i < n[u]; ++i) {if (s[u][i][0] <= 0 || s[u][i][1] > mn) {continue;}// 找第一个活着攻击次数最小的int hpu = s[u][i][0];int atku = atk[u][i];for (int j = 0; j < n[v]; ++j) {if (s[v][j][0] <= 0) {continue;}int hpv = s[v][j][0];int atkv = atk[v][j];s[u][i][0] -= atkv;s[u][i][1] += 1;s[v][j][0] -= atku;auto ret = dfs(v);for (int k = 0; k < 3; ++k) {res[k] += ret[k] / cnt[v];}s[u][i][0] += atkv;s[u][i][1] -= 1;s[v][j][0] += atku;}break;}return res;
}std::array<long double, 3> dfs() {std::array<long double, 3> res = { 0.0, 0.0, 0.0 };auto ret0 = dfs(0);auto ret1 = dfs(1);for (int i = 0; i < 3; ++i) {res[i] = (ret0[i] + ret1[i]) / 2.0;}return res;
}void solve() {std::cin >> n[0] >> n[1];for (int i = 0; i < 2; ++i) {for (int j = 0; j < n[i]; ++j) {std::cin >> atk[i][j];s[i][j] = { atk[i][j], 0 };}}std::cout << std::fixed << std::setprecision(10);if (n[0] > n[1]) {for (auto &i : dfs(0)) {std::cout << i << '\n';}}else if (n[1] > n[0]) {for (auto &i : dfs(1)) {std::cout << i << '\n';}}else {for (auto &i : dfs()) {std::cout << i << '\n';}}
}