2016 ACM/ICPC亚洲区沈阳站
| A | B | C | D | E | F | G | H | I | J | K | L | M |
| O | O | O | $\varnothing$ | $\varnothing$ | $\varnothing$ | $\varnothing$ | $\varnothing$ | $\varnothing$ | $\varnothing$ |
签到。
签到
$$f[i] = f[i - 1] + 2 * f[i - 2] + i ^ 4$$
$$
\left[
\begin{matrix}
1 & 2 & 1 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 &0 & 1 & 4 & 6 &4 & 1 \\
0 & 0 & 0 & 1 & 3 & 3 & 1 \\
0 & 0 & 0 & 0 & 1 & 2 & 1 \\
0 & 0 & 0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1
\end{matrix}
\right]
\times
\left[
\begin{matrix}
f_{i-1} & 0 & 0 & 0 & 0 & 0 & 0 \\
f_{i-2} & 0 & 0 & 0 & 0 & 0 & 0\\
i^4 &0 & 0 & 0& 0 &0 & 0 \\
i^3 & 0 & 0 & 0 & 0 & 0 & 0 \\
i^2 & 0 & 0 & 0 & 0 & 0 & 0\\
i & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0
\end{matrix}
\right]=
\left[
\begin{matrix}
f_{i} & 0 & 0 & 0 & 0 & 0 & 0 \\
f_{i} & 0 & 0 & 0 & 0 & 0 & 0\\
(i+1)^4 &0 & 0 & 0& 0 &0 & 0 \\
(i+1)^3 & 0 & 0 & 0 & 0 & 0 & 0 \\
(i+1)^2 & 0 & 0 & 0 & 0 & 0 & 0\\
i+1 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0
\end{matrix}
\right]
$$
#include <bits/stdc++.h> #define ll long long const ll MOD = 2147493647; const int N = 7; struct Mat { ll a[10][10]; Mat() { memset(a, 0, sizeof(a)); } Mat(int x) { memset(a, 0, sizeof(a)); for (int i = 1; i <= N; i++) a[i][i] = 1; } Mat operator * (const Mat &rhs) const { Mat c; for (int i = 1; i <= N; i++) for (int j = 1; j <= N; j++) for (int k = 1; k <= N; k++) c.a[i][j] = (c.a[i][j] + a[i][k] * rhs.a[k][j] % MOD) % MOD; return c; } }; Mat qp(Mat a, int b) { Mat c(1); while (b) { if (b & 1) c = a * c; a = a * a; b >>= 1; } return c; } int main() { int T; scanf("%d", &T); Mat base; base.a[1][1] = 1; base.a[1][2] = 2; base.a[1][3] = 1; base.a[2][1] = 1; base.a[3][3] = 1; base.a[3][4] = 4; base.a[3][5] = 6; base.a[3][6] = 4; base.a[3][7] = 1; base.a[4][4] = 1; base.a[4][5] = 3; base.a[4][6] = 3; base.a[4][7] = 1; base.a[5][5] = 1; base.a[5][6] = 2; base.a[5][7] = 1; base.a[6][6] = 1; base.a[6][7] = 1; base.a[7][7] = 1; while (T--) { int n, a, b; scanf("%d%d%d", &n, &a, &b); if (n == 1) { printf("%d\n", a); continue; } if (n == 2) { printf("%d\n", b); continue; } n--; n--; Mat ans; ans.a[1][1] = b; ans.a[2][1] = a; ans.a[3][1] = 81; ans.a[4][1] = 27; ans.a[5][1] = 9; ans.a[6][1] = 3; ans.a[7][1] = 1; ans = qp(base, n) * ans; printf("%lld\n", ans.a[1][1] % MOD); } return 0; }
博弈。
$dp[n][a][b]$ 表示当前剩下 $n$ 个物品，第一个人有 $a$ 元钱，第二个人有 $b$ 元钱，第一个人能获得的物品数。
$dp[1][a][b] = [a\leq b]$
消除奇偶性可以通过从 $dp[i - 1][b][a]$ 转移过来。这样就不用考虑奇偶了。
然后枚举两个人分别要出多少钱，当第一个人出 $x$ 块钱，第二个人出 $x+1$ 块钱会使第一个人的收益变小，那么第二个人会继续加价。同理第一个人会继续加价。当无法得到更好的收益的时候就停下来。
#include <cstdio> #include <algorithm> #include <cstring> const int N = 256; short dp[N][N][N]; int n, T, a, b; inline int geta(int n, int a) { return n / 2 * a + (n - n / 2) * (a + 1); } int main() { for (int a = 0; a < N; a++) for (int b = 0; b < N; b++) if (a >= b) dp[1][a][b] = 1; for (int i = 2; i < N; i++) { for (int a = 0; a < N; a++) { int limit = std::min(N, geta(i, a)); for (int b = 0; b < limit; b++) { if (a == b) { dp[i][a][b] = (i + 1) / 2; continue; } int vala = i - dp[i - 1][b][a]; int valb = 0; for (int u = 1; ; u++) { if (b < u || (valb = (i - 1 - dp[i - 1][b - u][a])) >= vala) { dp[i][a][b] = vala; break; } if (a < u || (vala = (i - dp[i - 1][b][a - u])) <= valb) { dp[i][a][b] = valb; break; } } } } } scanf("%d", &T); while (T--) { scanf("%d%d%d", &n, &a, &b); int ans1 = dp[n][a][b], ans2 = n - ans1; printf("Alice %d Bob %d\n", ans1, ans2); } return 0; }
经过暑假牛客多校的洗礼，看到团就想到暴搜...
首先用了bfs+bitset。T了。
改成vector判，又T了。
看了一份题解是dfs，存团的是数组。
我把那份代码的数组改成vector，还是T。
这么卡STL的吗...
#include <bits/stdc++.h> const int N = 107; std::vector<int> G[N]; bool mp[N][N]; int ans, n, m, s; void solve(int *a, int u) { if (a[0] == s) { ans++; return; } for (int v: G[u]) { if (v <= u) continue; bool flag = 1; for (int j = 1; j <= a[0]; j++) { if (!mp[a[j]][v]) { flag = 0; break; } } if (!flag) continue; a[++a[0]] = v; solve(a, v); --a[0]; } } int a[N]; int main() { int T; scanf("%d", &T); while (T--) { scanf("%d%d%d", &n, &m, &s); for (int i = 0; i <= n; i++) { G[i].clear(); for (int j = 0; j <= n; j++) mp[i][j] = 0; a[i] = 0; } for (int i = 0; i < m; i++) { int u, v; scanf("%d%d", &u, &v); if (u > v) std::swap(u, v); G[u].push_back(v); mp[u][v] = mp[v][u] = 1; } ans = 0; for (int i = 1; i <= n; i++) { a[0] = 1; a[1] = i; solve(a, i); } printf("%d\n", ans); } return 0; }
积分题。
看的这篇 https://www.cnblogs.com/chen9510/p/7635679.html
#include <bits/stdc++.h> const double pi = acos(-1.0); const double eps = 1e-10; inline int dcmp(double x) { if (fabs(x) < eps) return 0; return x < 0 ? -1 : 1; } double cal(double x) { return sin(x) - x * cos(x) - 1 / 3.0 * sin(x) * sin(x) * sin(x); } double calV(double mid) { double V = cal(acos(1.0)) - cal(acos(1.0 - mid)); V *= -2.0 / mid; return V; } double cal2(double x) { return x + sin(2 * x) / 2; } double area(double a, double x) { return (cal2(pi / 2.0) - cal2(asin(x / a))) * a; } int main() { int T; scanf("%d", &T); while (T--) { double d; scanf("%lf", &d); if (dcmp(d - 1) >= 0) { double h = 4 - 2 * d; double a = sqrt(2 * 2 + h * h) / 2; printf("%.5f\n", pi * a); continue; } if (dcmp(d) == 0) { puts("0.00000"); continue; } double l = 0, r = 2.0; for (int i = 0; i < 50; i++) { double mid = (l + r) / 2.0; double V = calV(mid); if (dcmp(V - pi * d) == 0) break; if (dcmp(V - pi * d) < 0) l = mid; else r = mid; } double mid = l; int flag = 1; if (mid < 1) flag = -1; double len = sqrt(mid * mid + 4); double h = sqrt(1 - (1 - mid) * (1 - mid)); double a = len / (1 + flag * sqrt(1 - h * h)); double x = a - len; printf("%.5f\n", area(a, x)); } }
先把AC自动机建出来。在AC自动机上DP。
$dp[i] = \sum \frac{1}{6} \times dp[from]$
$from$ 为能转到 $i$ 且不为某个串的结尾的结点
因为会存在环，所以高斯消元就行了。
#include <bits/stdc++.h> const int N = 100 + 7; double mat[N][N]; const double eps = 1e-8; inline void gauss(int n) { for(int i = 0; i <= n; i++) { int r = i; for(int j = i + 1; j <= n; j++) if(std::fabs(mat[r][i]) < std::fabs(mat[j][i])) r = j; if(r != i) std::swap(mat[i], mat[r]); for(int j = 0; j <= n; j++) { if(j == i) continue; double t = mat[j][i] / mat[i][i]; for(int k = i; k <= n + 1; k++) mat[j][k] -= mat[i][k] * t; } } for(int i = 1; i <= n; i++) { mat[i][n + 1] /= mat[i][i]; } } struct Aho { static const int sz = 6; int ch[N][sz], last[N], fail[N], tol; bool end[N]; void init() { tol = 0; newnode(); } inline int newnode() { memset(ch[tol], 0, sizeof(ch[tol])); last[tol] = fail[tol] = end[tol] = 0; return tol++; } void insert(int *a, int n) { int u = 0; for (int i = 0; i < n; i++) { int id = a[i] - 1; if (!ch[u][id]) ch[u][id] = newnode(); u = ch[u][id]; } end[u] = 1; } void build() { std::queue<int> que; for (int i = 0; i < sz; i++) if (ch[0][i]) que.push(ch[0][i]), fail[ch[0][i]] = last[ch[0][i]] = 0; while (!que.empty()) { int u = que.front(); que.pop(); end[u] |= end[last[u]]; for (int i = 0; i < sz; i++) { int &v = ch[u][i]; if (v) { fail[v] = ch[fail[u]][i]; que.push(v); last[v] = end[fail[v]] ? fail[v] : last[fail[v]]; } else { v = ch[fail[u]][i]; } } } } void solve() { memset(mat, 0, sizeof(mat)); mat[0][tol] = -1.0; for (int i = 0; i < tol; i++) { mat[i][i] = -1.0; if (end[i]) continue; for (int j = 0; j < sz; j++) mat[ch[i][j]][i] += 1.0 / 6; } gauss(tol - 1); bool flag = 0; for (int i = 0; i < tol; i++) if (end[i]) { if (flag) putchar(' '); printf("%.6f", mat[i][tol]); flag = 1; } puts(""); } } ac; int a[20]; int main() { int T; scanf("%d", &T); for (; T--; ) { int n, l; scanf("%d%d", &n, &l); ac.init(); for (int i = 1; i <= n; i++) { for (int j = 0; j < l; j++) scanf("%d", a + j); ac.insert(a, l); } ac.build(); ac.solve(); } return 0; }
$dp[u] = min(dp[anc] + (sum[u] - sum[anc])^2 + p)$
$anc$ 为 $u$ 到根的路径上的结点。
斜率优化DP一下。在进入一个结点时存储一下对当前队列的修改，离开一个结点时改回去，这样就能保证进入一个结点时，队列存储的都是它的祖先。
#include <bits/stdc++.h> #define pii pair<int, int> #define ll long long #define fi first #define se second const int N = 1e5 + 7; const double eps = 1e-10; ll dp[N]; int n, p, que[N], l, r; ll sum[N]; std::vector<std::pii> G[N]; inline ll sqr(ll x) { return x * x; } inline ll up(int i, int j) { return dp[i] + sqr(sum[i]) - dp[j] - sqr(sum[j]); } inline ll down(int i, int j) { return sum[i] - sum[j]; } ll ans; void dfs(int u, int fa = 0) { std::vector<std::pii> vec; int x = l, y = r; while (l < r && up(que[l + 1], que[l]) <= down(que[l + 1], que[l]) * 2 * sum[u]) { vec.push_back(std::pii(l, que[l])); l++; } if (u != 1) { dp[u] = dp[que[l]] + sqr(sum[u] - sum[que[l]]) + p; ans = std::max(ans, dp[u]); } while (l < r && up(que[r], que[r - 1]) * down(u, que[r]) >= up(u, que[r]) * down(que[r], que[r - 1])) { vec.push_back(std::pii(r, que[r])); r--; } que[++r] = u; for (auto v: G[u]) { if (v.fi == fa) continue; sum[v.fi] = sum[u] + v.se; dfs(v.fi, u); } l = x, r = y; for (auto p: vec) que[p.fi] = p.se; } int main() { int T; scanf("%d", &T); while (T--) { scanf("%d%d", &n, &p); for (int i = 1; i <= n; i++) dp[i] = sum[i] = 0, G[i].clear(); for (int i = 1, u, v, w; i < n; i++) { scanf("%d%d%d", &u, &v, &w); G[u].push_back(std::pii(v, w)); G[v].push_back(std::pii(u, w)); } dp[1] = -p; que[l = r = 1] = 0; ans = 0; dfs(1); printf("%lld\n", ans); } return 0; }
基环树先找出环，用一个数组 $a$ 记录位置。然后以环上的每个点为根，去bfs非环上的点得到bfs序.
那么对于非环上的点，与他距离不大于 $k$ 的点bfs序连续，对于环上的点，与他距离不大于 $k$ 的点在 $a$ 数组里连续
这样就可以用线段树维护了。
然后就是恶心的细节了。
#include <bits/stdc++.h> #define ll long long inline void checkmax(int &a, int b) { if (a < b) a = b; } inline void checkmin(int &a, int b) { if (a > b) a = b; } const int N = 1e5 + 7; std::vector<int> vec[N]; int fa[N], son[N], n, id[N], BCC[N], a[N]; int tol; bool vis[N]; void getBCC(int u, int v) { for (int i = u; i != v; i = fa[i]) a[BCC[i] = ++a[0]] = i; a[BCC[v] = ++a[0]] = v; } void dfs(int u, int pre) { vis[u] = 1; for (int v: vec[u]) { if (v == pre) continue; if (!vis[v]) { fa[v] = u; dfs(v, u); } else if (!BCC[u]) { getBCC(u, v); } } } int que[N]; int ls[N], rs[N], lg[N], rg[N]; void bfs(int s) { int h = 0, t = 0; que[++t] = s; id[s] = ++tol; while (h != t) { int u = que[++h]; ls[u] = tol + 1; for (int v: vec[u]) if (!id[v] && !BCC[v]) { fa[v] = u; que[++t] = v; id[v] = ++tol; } rs[u] = tol; } for (int i = 1; i <= t; i++) { int u = que[i]; lg[u] = N; rg[u] = -N; for (int v: vec[u]) if (id[v] > id[u] && !BCC[v]) checkmin(lg[u], ls[v]), checkmax(rg[u], rs[v]); } } struct Seg { #define lp p << 1 #define rp p << 1 | 1 static const int NN = N * 4; ll sum[NN], lazy[NN]; void build(int p, int l, int r) { sum[p] = lazy[p] = 0; if (l == r) return; int mid = l + r >> 1; build(lp, l, mid); build(rp, mid + 1, r); } inline void pushup(int p) { sum[p] = sum[lp] + sum[rp]; } inline void tag(int p, int len, ll w) { lazy[p] += w; sum[p] += len * w; } inline void pushdown(int p, int llen, int rlen) { if (lazy[p]) { tag(lp, llen, lazy[p]); tag(rp, rlen, lazy[p]); lazy[p] = 0; } } void update(int p, int l, int r, int x, int y, int w) { if (x > r || l > y) return; if (x <= l && y >= r) { tag(p, r - l + 1, w); return; } int mid = l + r >> 1; pushdown(p, mid - l + 1, r - mid); if (x <= mid) update(lp, l, mid, x, y, w); if (y > mid) update(rp, mid + 1, r, x, y, w); pushup(p); } ll query(int p, int l, int r, int x, int y) { if (x > r || l > y) return 0; if (x <= l && y >= r) return sum[p]; int mid = l + r >> 1; pushdown(p, mid - l + 1, r - mid); ll ans = 0; if (x <= mid) ans += query(lp, l, mid, x, y); if (y > mid) ans += query(rp, mid + 1, r, x, y); return ans; } inline int getl(int x) { return (x > 1) ? x - 1 : a[0]; } inline int getr(int x) { return (x < a[0]) ? x + 1: 1; } inline void update(int u, int w) { update(1, 1, tol, u, u, w); } inline ll query(int u) { return query(1, 1, tol, u, u); } void update(int u, int k, int w) { update(id[u], w); if (k >= 1) { update(1, 1, tol, ls[u], rs[u], w); if (!BCC[u]) update(id[fa[u]], w); else { update(id[a[getl(BCC[u])]], w); update(id[a[getr(BCC[u])]], w); } } if (k >= 2) { update(1, 1, tol, lg[u], rg[u], w); if (!BCC[u]) { update(id[u], -w); update(1, 1, tol, ls[fa[u]], rs[fa[u]], w); int p = BCC[fa[u]]; if (!p) update(id[fa[fa[u]]], w); else update(id[a[getl(p)]], w), update(id[a[getr(p)]], w); } else { int bl = getl(BCC[u]), br = getr(BCC[u]); update(1, 1, tol, ls[a[bl]], rs[a[bl]], w); update(1, 1, tol, ls[a[br]], rs[a[br]], w); if (getl(bl) != br) update(id[a[getl(bl)]], w); if (getr(br) != bl && getr(br) != getl(bl)) update(id[a[getr(br)]], w); } } } ll query(int u, int k) { ll ans = query(id[u]); if (k >= 1) { ans += query(1, 1, tol, ls[u], rs[u]); if (!BCC[u]) ans += query(id[fa[u]]); else ans += query(id[a[getl(BCC[u])]]) + query(id[a[getr(BCC[u])]]); } if (k >= 2) { ans += query(1, 1, tol, lg[u], rg[u]); if (!BCC[u]) { ans += query(1, 1, tol, ls[fa[u]], rs[fa[u]]) - query(id[u]); int p = BCC[fa[u]]; if (!p) ans += query(id[fa[fa[u]]]); else ans += query(id[a[getl(p)]]) + query(id[a[getr(p)]]); } else { int bl = getl(BCC[u]), br = getr(BCC[u]); ans += query(1, 1, tol, ls[a[bl]], rs[a[bl]]) + query(1, 1, tol, ls[a[br]], rs[a[br]]); if (getl(bl) != br) ans += query(id[a[getl(bl)]]); if (getr(br) != bl && getr(br) != getl(bl)) ans += query(id[a[getr(br)]]); } } return ans; } } seg; int main() { freopen("in.txt", "r", stdin); int T; scanf("%d", &T); while (T--) { scanf("%d", &n); tol = a[0] = 0; for (int i = 1; i <= n; i++) { fa[i] = son[i] = BCC[i] = id[i] = vis[i] = 0; vec[i].clear(); } for (int i = 1, u, v; i <= n; i++) { scanf("%d%d", &u, &v); vec[u].push_back(v); vec[v].push_back(u); } dfs(1, 0); for (int i = 1; i <= a[0]; i++) bfs(a[i]); seg.build(1, 1, tol); int q; scanf("%d", &q); while (q--) { int u, k; char s[20]; scanf("%s%d%d", s, &u, &k); if (s[0] == 'M') { int d; scanf("%d", &d); seg.update(u, k, d); } else { printf("%lld\n", seg.query(u, k)); } } } return 0; }
终于把这道题给补了...比赛前我居然还在搞这些不考的东西
裸 K 短路，并且是有向无环图，求 $T$ 到其他点的最短路拓扑排序就能解决。
求出 $T$ 到其他所有点的最短路树，记 $d[i]$ 为 $i$ 到 $T$ 的最短路。给每一个点分配一个前趋，如果多个相同则选其中一个。（注意有重边时要记录边而不是记录前趋的点！！！）
走 $S$ 到 $T$ 上的树边即为最短路，走一条非树边 $(u, v, c)$ 则会使费用增大 $d[v] + c - d[u]$。记该花费为非树边的费用，显然树边的费用为 $0$。
将一条从 $S$ 到 $T$ 的路径记为其经过的非树边序列 $p$，那么这条路径的权值和即为 $d[s] + \sum_{(u,v,c) \in p} (d[v] + c - d[u])$
求 $k$ 短路即求第 $k$ 小的合法非树边序列费用之和。
合法的非树边序列为相邻两条非树边 $e$，$f$，$e$ 在 $f$ 之前，$head(f)$ 需为 $tail(e)$ 在 $T$ 上的祖先或相同。
用一个堆来存储搜索的状态，当前的非树边权值和为优先级，再存储最后一条非树边的起点。
往后可以有两个决策，一为直接加上最后一条非树边的终点之后的非树边中，权值最小的那个。
二为将最后一条非树边替换为 $u$ 之后的非树边下一条比这条非树边大的。
发现需要用另一个堆维护每个点往后所有的非树边。发现 $u$ 和 $pre[u]$ 大部分非树边相同，只是多了一些以自身为起点的非树边，那么可以可持久化地添加非树边。
可以用可持久化左偏树，虽然论文里说的是它不可完全可持久化。
还是挺好写的，就是在合并的过程中用新的节点来合并。像线段树合并。
#include <bits/stdc++.h> #define pii pair<int, int> #define fi first #define se second const int N = 4e5 + 7; const int INF = 0x3f3f3f3f; template<class T>inline bool chkmin(T &a, const T &b) { return a > b ? a = b, true : false; } template<class T>inline bool chkmax(T &a, const T &b) { return a < b ? a = b, true : false; } namespace Heap { struct Node { int lp, rp, v, dis, val; } tree[N * 20]; int tol, root[N]; int newnode(int x = 0, int y = 0) { int p = ++tol; tree[p].val = x, tree[p].v = y; tree[p].lp = tree[p].rp = 0; tree[p].dis = 1; return p; } int merge(int p, int q) { if (!p || !q) return p + q; if (tree[p].val < tree[q].val) std::swap(p, q); int x = newnode(); tree[x] = tree[p]; tree[x].rp = merge(tree[x].rp, q); if (tree[tree[x].lp].dis < tree[tree[x].rp].dis) std::swap(tree[x].lp, tree[x].rp); tree[x].dis = tree[tree[x].rp].dis + 1; return x; } inline void init() { tol = 0; tree[0].dis = -1; } } using namespace Heap; int n, k, cnt, id[N][2], s, t, deg[N]; int head1[N], head2[N], e, d[N]; struct Ed { int v, ne, c; } E[N]; void add(int u, int v, int c) { E[cnt].v = v; E[cnt].ne = head1[u]; E[cnt].c = c; head1[u] = cnt++; E[cnt].v = u; E[cnt].ne = head2[v]; E[cnt].c = c; head2[v] = cnt++; deg[u]++; } int Q[N], pre[N]; int topo(int S, int *head) { int l = 0, r = 0; Q[r++] = S; d[S] = 0; while (l <= r) { int u = Q[l++]; for (int i = head[u]; ~i; i = E[i].ne) { int v = E[i].v; if (chkmax(d[v], d[u] + E[i].c)) pre[v] = i; if (!--deg[v]) Q[r++] = v; } } return r; } int solve() { scanf("%d%d", &n, &k); cnt = e = 0; init(); for (int i = 0; i <= n; i++) id[i][0] = ++e, id[i][1] = ++e; s = ++e, t = ++e; for (int i = 1; i <= e; i++) { root[i] = 0; d[i] = -INF; pre[i] = -1; deg[i] = 0; head1[i] = head2[i] = -1; } for (int i = 0; i < n; i++) { int a, b, c; scanf("%d%d%d", &a, &b, &c); add(id[i][0], id[i + 1][c], a); add(id[i][1], id[i + 1][c], b); add(id[i][0], id[i + 1][0], 0); add(id[i][1], id[i + 1][1], 0); } add(s, id[0][0], 0); add(id[n][0], t, 0); add(id[n][1], t, 0); int m = topo(t, head2); if (k == 1) return d[s]; for (int j = 0; j < m; j++) { int u = Q[j]; for (int i = head1[u]; ~i; i = E[i].ne) { int v = E[i].v; if (d[v] == INF || (i ^ 1) == pre[u]) continue; root[u] = merge(root[u], newnode(d[v] + E[i].c - d[u], v)); } if (~pre[u]) root[u] = merge(root[u], root[E[pre[u] ^ 1].v]); } std::priority_queue<std::pii> que; que.push(std::pii(d[s] + tree[root[s]].val, root[s])); for (k -= 2; k; k--) { auto p = que.top(); que.pop(); int i = p.se, j = root[tree[i].v]; if (j) que.push(std::pii(p.fi + tree[j].val, j)); if (tree[i].lp) que.push(std::pii(p.fi - tree[i].val + tree[tree[i].lp].val, tree[i].lp)); if (tree[i].rp) que.push(std::pii(p.fi - tree[i].val + tree[tree[i].rp].val, tree[i].rp)); } return que.top().fi; } int main() { freopen("in.txt", "r", stdin); int T; scanf("%d", &T); while (T--) printf("%d\n", solve()); return 0; }
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