牛客多校1-H Longest Path

题意一棵树，两点的路径和d(u,v)为相邻边权差的平方和的总和。对于每个点，求$max_vd(u,v)$

思路 经典树型DP+斜率优化

我们考虑每个点向上走与向下走能获得的答案，那么每个点总的答案就是$max(up[i], down[i])$ 向下走的答案是容易求得的，只需要从叶子向上DP即可。向上走的答案就要麻烦一些了。对于一个点

$up[i] = max(up[i_{father}] + (c[i] - c[i_{father}])^2, down[i_{brother}] + (c[i] - c[i_{brother}])^2)$

这里出现了二次项，是一个经典的斜率优化DP。考虑

$max\lbrace down[i_{brother}] + (c[i] - c[i_{brother}])^2 \rbrace$

把这个式子拆开来， 我们要从j转移而不是从k转移的条件是

$(down[j] + c[j]^2) - (down[k] + c[k]^2) > 2*c[i]*(c[j] - c[k])$

对于每个状态，将其看作一个$point(c[j], c[j]^2 + down[j])$ 则该式子可看作$c[i]$ <= $p_j$与$p_k$ 两点的斜率($c[j] > c[k]$的情况下), up[i]将从j转移过来。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;

const int N = 1e5 + 10;
ll sqr(ll x){
    return x*x;
}
struct point{
    ll x,y;
    int id;
    bool operator < (point &b){
        return x < b.x;
    }
    point operator - (point &p){
        return {x - p.x, y - p.y};
    }
};
ll dot(point a, point b) {
    return a.x * b.x + a.y * b.y;
}
ll cross(point a, point b) {
    return a.x * b.y - a.y * b.x;
}
vector<point>G[N];
ll down[N], up[N], ans[N], c[N];

void dfs1(int u, int fa){
    for (point e : G[u]){
        int v = e.x, w = e.y;
        if (v == fa) continue;
        c[v] = w;
        dfs1(v, u);
        if (fa) {
            down[u] = max(down[u], down[v] + sqr(c[v] - c[u]));
            ans[fa] = max(ans[fa], down[u]);
        }
    }
}

point ch[N], son[N];
int top, m;
void dfs2(int u, int fa){
    m = 0;
    for (point e : G[u]){
        int v = e.x;
        if (v == fa) continue;
        if (fa) up[v] = max(up[v], up[u] + sqr(c[v] - c[u]));
        son[m++] = {c[v], down[v] + sqr(c[v]), v};
    }
    sort(son, son + m);
    top=0;
    for (int i = 0; i < m; i++){
        int id = son[i].id;
        point vec = {-2 * c[id], 1};
        while(top > 1 && dot(vec, ch[top - 1]) <= dot(vec, ch[top - 2])) top--;
        if (top) up[id] = max(up[id], dot(vec, ch[top - 1]) + sqr(c[id]));
        while(top > 1 && cross(ch[top - 1] - ch[top - 2], son[i] - ch[top - 2]) >= 0) top--;
        ch[top++] = son[i];
    }
    top = 0;
    for (int i = m - 1; i >= 0; i--){
        int id = son[i].id;
        point vec={-2 * c[id], 1};
        while(top > 1 && dot(vec, ch[top - 1]) <= dot(vec, ch[top - 2])) top--;
        if (top) up[id] = max(up[id], dot(vec, ch[top - 1]) + sqr(c[id]));
        while(top > 1 && cross(ch[top - 1] - ch[top - 2], son[i] - ch[top - 2]) <= 0) top--;
        ch[top++] = son[i];
    }
    for (point e : G[u]){
        int v = e.x;
        if (v == fa) continue;
        dfs2(v, u);
    }
}
int main(){
    int n;
    while(~scanf("%d", &n)){
        for (int i = 1;i <= n; i++) {
            G[i].clear();
            ans[i] = down[i] = up[i] = c[i] = 0;
        }
        for (int i = 1, x, y, z; i < n; i++){
            scanf("%d%d%d", &x, &y, &z);
            G[x].push_back({y, z});
            G[y].push_back({x, z});
        }
        dfs1(1, 0);
        dfs2(1, 0);
        for (int i = 1;i <= n; i++) printf("%lld\n", max(ans[i], up[i]));
    }
}