「CTSC2008」网络管理

「CTSC2008」网络管理

传送门

整体二分做法，应该和这题一样的吧。

就是把序列换成树，第 \(k\) 小换成第 \(k\) 大。

然后就切了。。。

参考代码：

#include <algorithm>
#include <cstdio>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
using namespace std;
template < class T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while ('0' > c || c > '9') f |= c == '-', c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}

const int _ = 80005;

int tot, head[_]; struct Edge { int ver, nxt; } edge[_ << 1];
inline void Add_edge(int u, int v) { edge[++tot] = (Edge) { v, head[u] }, head[u] = tot; }

int n, m, a[_], ans[_], tr[_];
int num, q; struct node { int k, x, y, id; } t[_ << 1], tt1[_ << 1], tt2[_ << 1];
int dep[_], siz[_], son[_], fa[_], dfn[_], top[_];

inline void update(int x, int v) { for (rg int i = x; i <= n; i += i & -i) tr[i] += v; }

inline int query(int x) { int res = 0; for (rg int i = x; i >= 1; i -= i & -i) res += tr[i]; return res; }

inline int Query(int x, int y) {
    int fx = top[x], fy = top[y], res = 0;
    while (fx != fy) {
    	if (dep[fx] < dep[fy]) swap(x, y), swap(fx, fy);
    	res += query(dfn[x]) - query(dfn[fx] - 1), x = fa[fx], fx = top[x];
    }
    if (dep[x] > dep[y]) swap(x, y);
    res += query(dfn[y]) - query(dfn[x] - 1);
    return res;
}

inline void dfs(int u, int f) {
    dep[u] = dep[f] + 1, siz[u] = 1, fa[u] = f;
    for (rg int i = head[u]; i; i = edge[i].nxt) {
    	int v = edge[i].ver; if (v == f) continue ;
    	dfs(v, u), siz[u] += siz[v];
    	if (siz[son[u]] < siz[v]) son[u] = v;
    }
}

inline void dfs(int u, int f, int topf) {
    dfn[u] = ++dfn[0], top[u] = topf;
    if (son[u]) dfs(son[u], u, topf);
    for (rg int i = head[u]; i; i = edge[i].nxt) {
	    int v = edge[i].ver; if (v == f || v == son[u]) continue ;
	    dfs(v, u, v);
    }
}

inline void solve(int ql, int qr, int l, int r) {
    if (ql > qr || l > r) return ;
    if (l == r) { for (rg int i = ql; i <= qr; ++i) if (t[i].id) ans[t[i].id] = l; return ; }
    int mid = (l + r) >> 1, p1 = 0, p2 = 0;
    for (rg int i = ql; i <= qr; ++i) {
	    if (t[i].id == 0) {
    	    if (t[i].y <= mid) tt1[++p1] = t[i]; else update(dfn[t[i].x], t[i].k), tt2[++p2] = t[i];
    	} else {
	        int cnt = Query(t[i].x, t[i].y);
	        if (cnt < t[i].k) t[i].k -= cnt, tt1[++p1] = t[i]; else tt2[++p2] = t[i];
	    }
    }
    for (rg int i = 1; i <= p2; ++i) if (tt2[i].id == 0) update(dfn[tt2[i].x], -tt2[i].k);
    for (rg int i = 1; i <= p1; ++i) t[ql + i - 1] = tt1[i];
    for (rg int i = 1; i <= p2; ++i) t[ql + p1 + i - 1] = tt2[i];
    solve(ql, ql + p1 - 1, l, mid), solve(ql + p1, qr, mid + 1, r);
}

int main() {
#ifndef ONLINE_JUDGE
    file("cpp");
#endif
    read(n), read(m);
    for (rg int i = 1; i <= n; ++i) read(a[i]), t[++num] = (node) { 1, i, a[i], 0 };
    for (rg int u, v, i = 1; i < n; ++i) read(u), read(v), Add_edge(u, v), Add_edge(v, u);
    dfs(1, 0), dfs(1, 0, 1);
    for (rg int k, x, y, i = 1; i <= m; ++i) {
	    read(k), read(x), read(y);
	    if (k == 0) {
    	    t[++num] = (node) { -1, x, a[x], 0 };
	        t[++num] = (node) { 1, x, a[x] = y, 0 };
	    } else
    	    t[++num] = (node) { k, x, y, ++q };
    }
    solve(1, num, 0, 100000000);
    for (rg int i = 1; i <= q; ++i) if (ans[i] == 0) puts("invalid request!"); else printf("%d\n", ans[i]);
    return 0;
}