P4292 [WC2010]重建计划

发表于 | 阅读数

树上求一条链，求

最大。

显然分数规划，二分就是否存在链长符合要求$\sum e_i.v-mid\geq 0$

点分治可以做，常数有点大。

长链剖分，对一颗子树$u$来说$f[v][i]+f[u][j]+e.v\geq 0,L\leq i+j+1\leq U$。 即可

为了避免重复，选择边插入边计算，然后一遍合并，但是这样是$n^2$，可以选择用线段树维护$dfn[v]+i，dfn[u]+j$。

  • 在长链转移的时候，询问最小值,线段树也容易操作。
  • 对比使用指针，更加明确操作的数组在哪，更好维护。$dp[dfn[x]+y]\rightarrow dp[x][y]$
  • 由于边权可能$\leq -10^6$，线段树以及$-INF$，注意开大。

$BZOJ$过得去，$luogu,TLE,90$

代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, double>
#define mk make_pair
const double INF = 1e18;
const int N = 1e6 + 10;
const int mod = INF + 7;
const double eps = 1e-4;
int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}
const int SEN = 1e5 + 10;
struct segmentTree
{
    double maxx[SEN << 2], lazy[SEN << 2];
    void addtag(int pos, int l, int r, double w)
    {
        lazy[pos] += w;
        maxx[pos] += w;
    }
    void pushdown(int pos, int l, int r)
    {
        if (lazy[pos] != 0)
        {
            double w = lazy[pos];
            int mid = (l + r) >> 1;
            addtag(pos << 1, l, mid, w);
            addtag(pos << 1 | 1, mid + 1, r, w);
            lazy[pos] = 0;
        }
    }
    void pushup(int pos)
    {
        maxx[pos] = max(maxx[pos << 1 | 1], maxx[pos << 1]);
    }
    double query(int ql, int qr, int pos, int l, int r)
    {
        if (ql > qr || ql <= 0 || qr <= 0)
            return -INF;
        if (ql <= l && r <= qr)
        {
            return maxx[pos];
        }
        pushdown(pos, l, r);
        int mid = (l + r) >> 1;
        double ans = -INF;
        if (ql <= mid)
            ans = max(ans, query(ql, qr, pos << 1, l, mid));
        if (qr > mid)
            ans = max(ans, query(ql, qr, pos << 1 | 1, mid + 1, r));
        return ans;
    }
    void update(int ql, int qr, double w, int pos, int l, int r)
    {
        if (ql <= l && r <= qr)
        {
            addtag(pos, l, r, w);
            return;
        }
        pushdown(pos, l, r);
        int mid = (l + r) >> 1;

        if (ql <= mid)
            update(ql, qr, w, pos << 1, l, mid);
        if (qr > mid)
            update(ql, qr, w, pos << 1 | 1, mid + 1, r);
        pushup(pos);
    }
    void modify(int k, double w, int pos, int l, int r)
    {
        if (l == r)
        {
            maxx[pos] = max(w, maxx[pos]);
            lazy[pos] = 0;
            return;
        }
        pushdown(pos, l, r);
        int mid = (l + r) >> 1;

        if (k <= mid)
            modify(k, w, pos << 1, l, mid);
        if (k > mid)
            modify(k, w, pos << 1 | 1, mid + 1, r);
        pushup(pos);
    }
    void init(int pos, int l, int r)
    {
        lazy[pos] = 0;
        maxx[pos] = -INF;
        if (l == r)
        {
            return;
        }
        int mid = (l + r) >> 1;
        init(pos << 1, l, mid);
        init(pos << 1 | 1, mid + 1, r);
    }
} T;
int L, R;
int n;
bool Flag;
vector<pii> g[N];

double dp[N];
int dfn[N];
int len[N], son[N];
int tot;
void dfs1(int x, int fa)
{
    for (pii now : g[x])
    {
        int to = now.first;
        if (to == fa)
            continue;
        dfs1(to, x);
        son[x] = len[to] > len[son[x]] ? to : son[x];
    }
    len[x] = len[son[x]] + 1;
}
void dfs2(int x, int fa)
{
    dfn[x] = ++tot;
    if (son[x])
        dfs2(son[x], x);
    for (pii now : g[x])
    {
        int to = now.first;
        if (to == fa || son[x] == to)
            continue;
        dfs2(to, x);
    }
}
void solve(int x, int fa)
{
    if (Flag)
        return;

    for (pii now : g[x])
    {
        int to = now.first;
        if (son[x] == to)
        {
            solve(son[x], x);

            T.update(dfn[x] + 1, dfn[x] + len[x] - 1, now.second, 1, 1, n);
        }
    }
    T.modify(dfn[x], 0, 1, 1, n);
    double ans = -INF;
    for (pii now : g[x])
    {
        int to = now.first;
        if (to == fa || son[x] == to)
            continue;
        solve(to, x);
        if (Flag)
            return;
        for (int i = 0; i < len[to]; i++)
        {
            int l = max(L - 1 - i, 0), r = min(len[x] - 1, (R - 1 - i));

            ans = max(ans, T.query(dfn[to] + i, dfn[to] + i, 1, 1, n) + T.query(dfn[x] + l, dfn[x] + r, 1, 1, n) + now.second);
        }

        for (int i = 0; i < len[to]; i++)
            T.modify(dfn[x] + i + 1, T.query(dfn[to] + i, dfn[to] + i, 1, 1, n) + now.second, 1, 1, n);
    }

    int l = max(L, 0), r = min(len[x] - 1, R);

    ans = max(ans, T.query(dfn[x] + l, dfn[x] + r, 1, 1, n));
    if (ans >= 0)
        Flag = 1;
}

bool check(double x)
{
    for (int i = 1; i <= n; i++)
        for (int j = 0; j < g[i].size(); j++)
            g[i][j].second -= x;

    Flag = 0;
    T.init(1, 1, n);
    solve(1, 0);

    for (int i = 1; i <= n; i++)
        for (int j = 0; j < g[i].size(); j++)
            g[i][j].second += x;
    return Flag;
}
int main()
{
    n = read();
    L = read(), R = read();
    for (int i = 1; i < n; i++)
    {
        int u = read(), v = read(), w = read();
        g[u].push_back(mk(v, w));
        g[v].push_back(mk(u, w));
    }

    dfs1(1, 0);
    dfs2(1, 0);
    //cout << check(4) << endl;
    double l = 0, r = 1e6;
    while (l + eps < r)
    {
        double mid = (l + r) / 2;
        if (check(mid))
            l = mid;
        else
            r = mid;
    }

    printf("%.3f", l);
}