SPOJ GSS7 - Can you answer these queries VII

板的不能再板，链剖+线段树或者是LCT随便维护。

感觉唯一要注意的是跳链的时候要对$x$向上跳和$y$向上跳的情况分开讨论，而不能直接$swap$，因为只有两段接触的端点才能相互合并，而且每一次向上跳的线段要放在已经合并完成之后的左端。

最后输出答案的时候要注意这时候$x$和$y$合并好的树链上其实左端点都是在上面的，而$x$和$y$其实是左端点相接触的，所以最后合并的时候$x.lmax + y.lmax$可能成为最优答案而不是$x.rmax + y.lmax$。

一开始线段树写手残了……

时间复杂度$O(nlog^2n)$。

Code:

#include <cstdio>
#include <cstring>
using namespace std;

const int N = 1e5 + ;

int n, qn, tot = , head[N], a[N], w[N];
int dfsc = , id[N], siz[N], dep[N], son[N], fa[N], top[N]; 

struct Edge {
    int to, nxt;
} e[N << ];

inline void add(int from, int to) {
    e[++tot].to = to;
    e[tot].nxt = head[from];
    head[from] = tot;
}

inline void read(int &X) {
    X = ; char ch = ; int op = ;
    for(; ch > '' || ch < ''; ch = getchar())
        if(ch == '-') op = -;
    for(; ch >= '' && ch <= ''; ch = getchar())
        X = (X << ) + (X << ) + ch - ;
    X *= op;
}    

inline void swap(int &x, int &y) {
    int t = x; x = y; y = t;
}

inline int max(int x, int y) {
    return x > y ? x : y;
}

inline int max(int x, int y, int z) {
    return max(max(x, y), z);
}

void dfs1(int x, int fat, int depth) {
    dep[x] = depth, fa[x] = fat, siz[x] = ;
    int maxson = -;
    for(int i = head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        if(y == fat) continue;
        dfs1(y, x, depth + );

        siz[x] += siz[y];
        if(siz[y] > maxson)
            maxson = siz[y], son[x] = y;
    }
}

void dfs2(int x, int topf) {
    top[x] = topf, w[id[x] = ++dfsc] = a[x];
    if(!son[x]) return;
    dfs2(son[x], topf);
    for(int i = head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        if(y == fa[x] || y == son[x]) continue;
        dfs2(y, y);
    }
}    

struct Node {
    int lmax, rmax, sum, res, val;
    bool tag;

    inline void init() {
           lmax = rmax = sum = res = val = ;
        tag = ;
    }

};

inline Node merge(Node x, Node y) {
    Node ans; ans.init();
    ans.sum = x.sum + y.sum;
    ans.lmax = max(x.lmax, x.sum + y.lmax);
    ans.rmax = max(y.rmax, y.sum + x.rmax);
    ans.res = max(x.res, y.res, x.rmax + y.lmax);
    return ans;
}

namespace SegT {
    Node s[N << ];

    #define lc p << 1
    #define rc p << 1 | 1
    #define mid ((l + r) >> 1)
    #define lmax(p) s[p].lmax
    #define rmax(p) s[p].rmax
    #define sum(p) s[p].sum
    #define res(p) s[p].res
    #define tag(p) s[p].tag
    #define val(p) s[p].val

    inline void up(int p) {
        if(!p) return;
        lmax(p) = max(lmax(lc), sum(lc) + lmax(rc));
        rmax(p) = max(rmax(rc), sum(rc) + rmax(lc));
        sum(p) = sum(lc) + sum(rc);
        res(p) = max(res(lc), res(rc), rmax(lc) + lmax(rc));
    }

    inline void down(int p, int l, int r) {
        if(!tag(p)) return;
        sum(lc) = (mid - l + ) * val(p), sum(rc) = (r - mid) * val(p);
        res(lc) = lmax(lc) = rmax(lc) = max(, val(p) * (mid - l + ));
        res(rc) = lmax(rc) = rmax(rc) = max(, val(p) * (r - mid));
        val(lc) = val(rc) = val(p);
        tag(lc) = tag(rc) = , tag(p) = ;
    }

    void build(int p, int l, int r) {
        s[p].init();
        if(l == r) {
            sum(p) = w[l];
            lmax(p) = rmax(p) = res(p) = max(, w[l]);
            return;
        }

        build(lc, l, mid);
        build(rc, mid + , r);
        up(p);
    }

    void modify(int p, int l, int r, int x, int y, int v) {
        if(x <= l && y >= r) {
            sum(p) = (r - l + ) * v;
            lmax(p) = rmax(p) = res(p) = max(, sum(p));
            tag(p) = , val(p) = v;
            return;
        }

        down(p, l, r);
        if(x <= mid) modify(lc, l, mid, x, y, v);
        if(y > mid) modify(rc, mid + , r, x, y, v);
        up(p);
    }

/*    Node query(int p, int l, int r, int x, int y) {
        if(x <= l && y >= r) return s[p];

        down(p, l, r);

        if(y <= mid) return query(lc, l, mid, x, y);
        if(x > mid) return query(rc, mid + 1, r, x, y);

        Node res, ln = query(lc, l, mid, x, mid), rn = query(rc, mid + 1, y, mid + 1, y);
        res.init();

        res.sum = ln.sum + rn.sum;
        res.lmax = max(ln.lmax, ln.sum + rn.lmax);
        res.rmax = max(rn.rmax, rn.sum + ln.rmax);
        res.res = max(ln.res, rn.res, ln.rmax + rn.lmax);

        return res;
    }    */

    Node query(int p, int l, int r, int x, int y) {
        if(x <= l && y >= r) return s[p];

        down(p, l, r);

        Node ln, rn, ans;
        ln.init(), rn.init(), ans.init();

        if(x <= mid) ln = query(lc, l, mid, x, y);
        if(y > mid) rn = query(rc, mid + , r, x, y);

        ans = merge(ln, rn);
        return ans;
    }

} using namespace SegT;

inline void mTree(int x, int y, int v) {
    for(; top[x] != top[y]; ) {
        if(dep[top[x]] < dep[top[y]]) swap(x, y);
        modify(, , n, id[top[x]], id[x], v);
        x = fa[top[x]];
    }
    if(dep[x] > dep[y]) swap(x, y);
    modify(, , n, id[x], id[y], v);
}

inline void qTree(int x, int y) {
    Node ln, rn; ln.init(), rn.init();
    for(; top[x] != top[y]; ) {
        if(dep[top[x]] > dep[top[y]])
            ln = merge(query(, , n, id[top[x]], id[x]), ln), x = fa[top[x]];
        else
            rn = merge(query(, , n, id[top[y]], id[y]), rn), y = fa[top[y]];
    }
    if(dep[x] > dep[y])
        ln = merge(query(, , n, id[y], id[x]), ln);
    else
        rn = merge(query(, , n, id[x], id[y]), rn);

    printf("%d\n", max(ln.res, rn.res, ln.lmax + rn.lmax));
}

int main() {
    read(n);
    for(int i = ; i <= n; i++) read(a[i]);
    for(int x, y, i = ; i < n; i++) {
        read(x), read(y);
        add(x, y), add(y, x);
    }

    dfs1(, , ), dfs2(, );
    build(, , n);

    read(qn);
    for(int op, x, y; qn--; ) {
        read(op), read(x), read(y);
        if(op == ) {
            int v; read(v);
            mTree(x, y, v);
        } else qTree(x, y);
    }

    return ;
}