uoj198【CTSC2016】时空旅行

传送门：http://uoj.ac/problem/198

【题解】

首先y、z是没有用的。。

然后式子就是w = (x0-xi)^2+ci的最小值，化出来可以变成一个直线的形式。

然后我们可以用线段树维护dfs序上的每个点。

每个点维护经过这个点的所有直线（标记永久化），也就是维护上凸壳。

然后我们把询问按照x排序，每次决策点只会后移。所以复杂度就有保证啦！

真**难写

还有一个十分有趣的事实啊

我用一个号交完ac在另一个号再交就RE了啊。。。

不管了反正过了

# include <vector>
# include <stdio.h>
# include <string.h>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e5 + 10;
const int mod = 1e9+7;

# define RG register
# define ST static

int n, q, s[M], ps[M];
ll c0, ans[M];
vector<int> b[M];
struct planet {
    int x; ll c;
    planet() {}
    planet(int x, ll c) : x(x), c(c) {}
    friend bool operator < (planet a, planet b) {
        return a.x<b.x;
    }
}p[M];

struct line {
    ll k, b;
    line() {}
    line(ll k, ll b) : k(k), b(b) {}
    inline ll set(int x) {
        return k*(ll)x+b;
    }
};

struct quest {
    int s, x0, id;
    quest() {}
    quest(int s, int x0, int id) : s(s), x0(x0), id(id) {}
    friend bool operator < (quest a, quest b) {
        return a.x0<b.x0;
    }
}qu[M];

int head[M], nxt[M], to[M], tot, in[M], out[M], DFN;
inline void add(int u, int v) {
    ++tot; nxt[tot] = head[u]; head[u] = tot; to[tot] = v;
}

inline void dfs(int x) {
    in[x] = ++DFN;
    for (int i=head[x]; i; i=nxt[i]) dfs(to[i]);
    out[x] = DFN;
}

inline bool cmp(int x, int y) {
    return p[x].x<p[y].x;
}

inline bool cmp_b(int x, int y) {
    return in[x] < in[y];
}

namespace SMT {
    const int M = 2e6 + 10;
    vector<line> w[M];
    int lst[M];

    inline bool canout(line a, line b, line c) {
        if(b.k == c.k) return b.b >= c.b;
        return 1.0*(b.b-a.b)*(a.k-c.k) > 1.0*(c.b-a.b) * (a.k-b.k);
    }

    inline void change(int x, int l, int r, int L, int R, line add) {
        if(L <= l && r <= R) {
            if(w[x].size()) {
                while(w[x].size() > 1 && canout(w[x][w[x].size()-2], w[x][w[x].size()-1], add)) w[x].pop_back();
                w[x].push_back(add);
            } else w[x].push_back(add);
            return ;
        }
        int mid = l+r>>1;
        if(L <= mid) change(x<<1, l, mid, L, R, add);
        if(R > mid) change(x<<1|1, mid+1, r, L, R, add);
    }

    inline ll query(int x, int l, int r, int pos, int x0) {
        while(lst[x]+1 < w[x].size() && w[x][lst[x]].set(x0) > w[x][lst[x]+1].set(x0)) ++lst[x];
        ll cur = (w[x].size() ? w[x][lst[x]].set(x0) : 1e18);
        if(l == r) return cur;
        int mid = l+r>>1;
        if(pos <= mid) return min(cur, query(x<<1, l, mid, pos, x0));
        else return min(cur, query(x<<1|1, mid+1, r, pos, x0));
    }

    inline void change(int L, int R, line add) {
        if(L>R) return;
        change(1, 1, DFN, L, R, add);
    }
    inline ll query(int pos, int x0) {
        return query(1, 1, DFN, pos, x0);
    }
}

int main() {
    scanf("%d%d%lld", &n, &q, &c0);
    for (int i=1, op, fr, id, x, c; i<n; ++i) {
        ps[i] = i;
        scanf("%d%d%d", &op, &fr, &id);
        if(op == 0) {
            scanf("%d%*d%*d%lld", &x, &c);
            s[id] = i;
            p[id] = planet(x, c);
        } else b[id].push_back(i);
        add(fr, i);
    }
    dfs(0);
    sort(ps+1, ps+n, cmp);
    for (int i=1, t; i<n; ++i) {
        if(s[ps[i]]) {
            int id = ps[i], se = s[id];
            line L = line(-2ll*p[id].x, 1ll*p[id].x*p[id].x+p[id].c);
            if(b[id].size()) {
                sort(b[id].begin(), b[id].end(), cmp_b);
                t = b[id].size();
                SMT::change(in[se], in[b[id][0]]-1, L);
                SMT::change(out[b[id][t-1]]+1, out[se], L);
                for (int j=1; j<t; ++j)
                    SMT::change(out[b[id][j-1]]+1, in[b[id][j]]-1, L);
            } else SMT::change(in[se], out[se], L);
        }
    }
    for (int i=1; i<=q; ++i) {
        scanf("%d%d", &qu[i].s, &qu[i].x0);
        qu[i].id=i;
    }
    sort(qu+1, qu+q+1);
    for (int i=1; i<=q; ++i)
        ans[qu[i].id] = 1ll * qu[i].x0 * qu[i].x0 + min(c0, SMT::query(in[qu[i].s], qu[i].x0));
    for (int i=1; i<=q; ++i) printf("%lld\n", ans[i]);
    return 0;
}