• [NOI2014]购票——斜率优化+树链剖分+线段树


    建议到UOJ上去交

    题解

    一眼(DP),先把转移方程写出来
    (dp[i])为从点(i)出发到点(1)的最小费用,那么存在转移

    [f[i]=min{f[j]+(d[i]-d[j])p[i]}+q[i]=min{f[j]-d[j]p[i]}+d[i]*p[i]+q[i] ]

    这个式子看起来可以斜率优化啊,往下化几步,可以得到类似下面的东西:
    (d[j] < d[k]),则当(frac{dp[j]-dp[k]}{d[j]-d[k]}geqslant p[i])时从(j)转移更优,否则从(k)转移更优
    这表明我们只需要维护一个下凸壳,转移时二分一下就行了
    假设这个问题是在序列上且没有距离限制,你就已经可以(O(nlogn))(A)掉它了
    加上距离限制时,我们可以拿个线段树维护一下每一小段的凸壳,查询时取个最大值
    挪到树上时,只需要上个树剖
    托上面两个东西的福,复杂度也变成了(O(nlog^3n))[手动滑稽]
    然后就是码码码

    #include <bits/stdc++.h>
    
    using namespace std;
    
    #define MAXN 200000
    #define vi vector<int>
    #define pb push_back
    #define ll long long
    #define INF 0x7f7f7f7f7f7f7f7f
    #define LIM 17
    
    int n, t;
    vi G[MAXN + 5];
    int summit[MAXN + 5], f[20][MAXN + 5], fa[MAXN + 5], top[MAXN + 5], sz[MAXN + 5], hson[MAXN + 5], id[MAXN + 5], dfn[MAXN + 5], dfn_clk;
    ll d[MAXN + 5], s[MAXN + 5], p[MAXN + 5], q[MAXN + 5], l[MAXN + 5], dp[MAXN + 5];
    
    namespace HLD {
        void dfs1(int u) {
            sz[u] = 1;
            f[0][u] = fa[u];
            for (int i = 1; i <= LIM; ++i) f[i][u] = f[i - 1][f[i - 1][u]];
            for (auto v : G[u]) {
                if (v == fa[u]) continue;
                d[v] = d[u] + s[v];
                dfs1(v);
                sz[u] += sz[v];
                if (sz[v] > sz[hson[u]]) hson[u] = v;
            }
        }
    
        void dfs2(int u, int tp) {
            top[u] = tp;
            dfn[u] = ++dfn_clk;
            id[dfn_clk] = u;
            if (hson[u]) dfs2(hson[u], tp);
            for (auto v : G[u]) {
                if (v == fa[u] || v == hson[u]) continue;
                dfs2(v, v);
            }
        }
    }
    
    double slope(int x, int y) {
        return static_cast<double>(dp[y] - dp[x])/(d[y] - d[x]);
    }
    
    void pre() {
        for (int i = 2; i <= n; ++i) {
            int u = i;
            for (int j = LIM; ~j; --j)
                if (d[i] - d[f[j][u]] <= l[i])
                    u = f[j][u];
            summit[i] = !u ? 1 : u;
        }
    }
    
    namespace SegementTree {
        #define mid ((l + r) >> 1)
        #define lson (o << 1)
        #define rson (o << 1 | 1)
    
        vi ch[4 * MAXN + 5];
    
        void addPoint(vi &v, int x) { // 把点x扔进下凸壳
            while (v.size() >= 2 && slope(x, v[v.size() - 2]) < slope(v[v.size() - 1], v[v.size() - 2])) v.pop_back();
            v.push_back(x);
        }
    
        ll get(vi &v, ll p) { // 在凸壳上二分斜率
            if (v.size() == 1) return dp[v[0]] - d[v[0]] * p;
            ll ret = INF;
            int l = 0, r = v.size() - 1, m;
            double s1, s2;
            while (l <= r) {
                m = (l + r) >> 1;
                if (m == v.size() - 1) {
                    s1 = slope(v[m - 1], v[m]);
                    ret = min(ret, dp[v[m]] - d[v[m]] * p);
                    if (s1 <= p) l = m + 1;
                    else r = m - 1;
                }
                else if (!m) {
                    s2 = slope(v[m], v[m + 1]);
                    ret = min(ret, dp[v[m]] - d[v[m]] * p);
                    if (s2 <= p) l = m + 1;
                    else r = m - 1;
                }
                else {
                    s1 = slope(v[m - 1], v[m]);
                    s2 = slope(v[m], v[m + 1]);
                    ret = min(ret, dp[v[m]] - d[v[m]] * p);
                    if (s1 <= p && s2 <= p) l = m + 1;
                    else if(s1 <= p && s2 > p) return min(ret, dp[v[m]] - d[v[m]] * p);
                    else r = m - 1;
                }
            }
            return ret;
        }
    
        void insert(int o, int l, int r, int x, int u) { // 插入一个点
            addPoint(ch[o], u);
            if (l == r) return ;
            if (x <= mid) insert(lson, l, mid, x, u);
            else insert(rson, mid + 1, r, x, u);
        }
    
        ll query(int o, int l, int r, int L, int R, ll p) { // 在那一堆凸壳中找最小值
            if (L <= l && r <= R) return get(ch[o], p);
            ll ret = INF;
            if (L <= mid) ret = min(ret, query(lson, l, mid, L, R, p));
            if (R > mid) ret = min(ret, query(rson, mid + 1, r, L, R, p));
            return ret;
        }
    
        #undef mid
        #undef lson
        #undef rson
    }
    using namespace SegementTree;
    
    void update(int u) { // 求点u的DP值
        dp[u] = INF;
        int x = fa[u];
        while (d[top[x]] >= d[summit[u]]) {
            dp[u] = min(dp[u], query(1, 1, n, dfn[top[x]], dfn[x], p[u]) + d[u] * p[u] + q[u]);
            x = fa[top[x]];
        }
        if (d[x] >= d[summit[u]]) dp[u] = min(dp[u], query(1, 1, n, dfn[summit[u]], dfn[x], p[u]) + d[u] * p[u] + q[u]);
    }
    
    void work(int u) {
        if (u != 1) update(u);
        insert(1, 1, n, dfn[u], u);
        for (auto v : G[u]) {
            if (v == fa[u]) continue;
            work(v);
        }
    }
    
    int main() {
        scanf("%d%d", &n, &t);
        for (int i = 2; i <= n; ++i) {
            scanf("%d%lld%lld%lld%lld", &fa[i], &s[i], &p[i], &q[i], &l[i]);
            G[fa[i]].pb(i);
        }
        HLD::dfs1(1);
        HLD::dfs2(1, 1);
        d[0] = -1;
        pre();
        work(1);
        for (int i = 2; i <= n; ++i) printf("%lld
    ", dp[i]);
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/dummyummy/p/11107075.html
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