SPOJ 6779 Can you answer these queries VII

SPOJ_6779

    这个题目和GSS1的思想是一样的，只不过由于变成了树，所以可以先用树链剖分先将树划分成若干棵线段树再进行求解。

    另外我的代码写得确实比较挫，推荐一份效率比较高的用动态树写的代码：http://ideone.com/wA3Pr。

#include<stdio.h>
#include<string.h>
#define MAXD 100010
#define MAXM 200010
#define INF 0x3f3f3f3f
int N, a[MAXD], mc[4 * MAXD], lc[4 * MAXD], rc[4 * MAXD], to[4 * MAXD], sum[4 * MAXD];
int first[MAXD], e, next[MAXM], v[MAXM];
int q[MAXD], fa[MAXD], dep[MAXD], size[MAXD], son[MAXD], top[MAXD], w[MAXD], cnt;
void Swap(int &x, int &y)
{
    int t;
    t = x, x = y, y = t;
}
int Max(int x, int y)
{
    return x > y ? x : y;
}
void add(int x, int y)
{
    v[e] = y;
    next[e] = first[x], first[x] = e ++;
}
void pushdown(int cur, int x, int y)
{
    if(to[cur] != INF)
    {
        int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
        to[ls] = to[rs] = to[cur];
        sum[ls] = (mid - x + 1) * to[cur], sum[rs] = (y - mid) * to[cur];
        mc[ls] = lc[ls] = rc[ls] = to[cur] > 0 ? sum[ls] : to[cur];
        mc[rs] = lc[rs] = rc[rs] = to[cur] > 0 ? sum[rs] : to[cur];
        to[cur] = INF;
    }
}
void update(int cur, int x, int y)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
    sum[cur] = sum[ls] + sum[rs];
    mc[cur] = Max(Max(mc[ls], mc[rs]), rc[ls] + lc[rs]);
    lc[cur] = Max(lc[ls], sum[ls] + lc[rs]);
    rc[cur] = Max(rc[rs], sum[rs] + rc[ls]);
}
void refresh(int cur, int x, int y, int s, int t, int v)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
    if(x >= s && y <= t)
    {
        to[cur] = v;
        sum[cur] = v * (y - x + 1);
        mc[cur] = lc[cur] = rc[cur] = v > 0 ? sum[cur] : v;
        return ;
    }
    pushdown(cur, x, y);
    if(mid >= s)
        refresh(ls, x, mid, s, t, v);
    if(mid + 1 <= t)
        refresh(rs, mid + 1, y, s, t, v);
    update(cur, x, y);
}
int Search(int cur, int x, int y, int s, int t, int flag, int &ans, int &ts, int &tc, int tflag)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
    if(x >= s && y <= t)
    {
        ans = Max(ans, mc[cur]);
        tc = Max(tc, ts + (tflag == 0 ? lc[cur] : rc[cur]));
        ts += sum[cur];
        return flag == 0 ? lc[cur] : rc[cur];
    }
    pushdown(cur, x, y);
    if(mid >= t)
        return Search(ls, x, mid, s, t, 0, ans, ts, tc, tflag);
    else if(mid + 1 <= s)
        return Search(rs, mid + 1, y, s, t, 1, ans, ts, tc, tflag);
    else
    {
        int ln, rn;
        if(tflag == 0)
            ln = Search(ls, x, mid, s, t, 1, ans, ts, tc, tflag), rn = Search(rs, mid + 1, y, s, t, 0, ans, ts, tc, tflag);
        else
            rn = Search(rs, mid + 1, y, s, t, 0, ans, ts, tc, tflag), ln = Search(ls, x, mid, s, t, 1, ans, ts, tc, tflag);
        ans = Max(ans, ln + rn);
        if(flag == 0)
            return Max(lc[ls], sum[ls] + rn);
        else
            return Max(rc[rs], sum[rs] + ln);
    }
}
void prepare()
{
    int i, j, k, x, rear = 0;
    q[rear ++] = 1;
    dep[1] = 1, fa[1] = 0;
    for(i = 0; i < rear; i ++)
    {
        x = q[i];
        for(j = first[x]; j != -1; j = next[j])
            if(v[j] != fa[x])
            {
                dep[v[j]] = dep[x] + 1;
                fa[v[j]] = x;
                q[rear ++] = v[j];
            }
    }
    size[0] = 0;
    for(i = rear - 1; i >= 0; i --)
    {
        x = q[i];
        son[x] = 0, size[x] = 1;
        for(j = first[x]; j != -1; j = next[j])
            if(v[j] != fa[x])
            {
                size[x] += size[v[j]];
                if(size[v[j]] > size[son[x]])
                    son[x] = v[j];
            }
    }
    memset(top, 0, sizeof(top));
    cnt = 0;
    for(i = 0; i < rear; i ++)
    {
        x = q[i];
        if(!top[x])
        {
            j = x;
            while(j != 0)
                top[j] = x, w[j] = ++ cnt, j = son[j];
        }
    }
}
void init()
{
    int i, j, k, x, y;
    for(i = 1; i <= N; i ++)
        scanf("%d", &a[i]);
    memset(first, -1, sizeof(first));
    e = 0;
    for(i = 1; i < N; i ++)
    {
        scanf("%d%d", &x, &y);
        add(x, y), add(y, x);
    }
    prepare();
    memset(to, 0x3f, sizeof(to));
    for(i = 1; i <= N; i ++)
        refresh(1, 1, N, w[i], w[i], a[i]);
}
void query(int x, int y)
{
    int fx = top[x], fy = top[y], cx, cy, ans, ts, tc;
    cx = cy = ans = 0;
    while(fx != fy)
    {
        if(dep[fx] > dep[fy])
            Swap(fx, fy), Swap(x, y), Swap(cx, cy);
        ts = tc = 0;
        Search(1, 1, N, w[fy], w[y], 0, ans, ts, tc, 1);
        ans = Max(ans, tc + cy);
        ts = tc = 0;
        Search(1, 1, N, w[fy], w[y], 0, ans, ts, tc, 0);
        cy = Max(ts + cy, tc);
        y = fa[fy], fy = top[y];
    }
    if(dep[x] > dep[y])
        Swap(x, y), Swap(cx, cy);
    ts = tc = 0;
    Search(1, 1, N, w[x], w[y], 0, ans, ts, tc, 0);
    ans = Max(ans, tc + cx);
    ts = tc = 0;
    Search(1, 1, N, w[x], w[y], 0, ans, ts, tc, 1);
    ans = Max(ans, tc + cy);
    ans = Max(ans, cx + ts + cy);
    printf("%d\n", ans);
}
void change(int x, int y, int v)
{
    int fx = top[x], fy = top[y];
    while(fx != fy)
    {
        if(dep[fx] > dep[fy])
            Swap(x, y), Swap(fx, fy);
        refresh(1, 1, N, w[fy], w[y], v);
        y = fa[fy], fy = top[y];
    }
    if(dep[x] > dep[y])
        Swap(x, y);
    refresh(1, 1, N, w[x], w[y], v);
}
void solve()
{
    int i, j, a, b, c, op, q;
    scanf("%d", &q);
    for(i = 0; i < q; i ++)
    {
        scanf("%d%d%d", &op, &a, &b);
        if(op == 1)
            query(a, b);
        else
        {
            scanf("%d", &c);
            change(a, b, c);
        }
    }
}
int main()
{
    while(scanf("%d", &N) == 1)
    {
        init();
        solve();
    }
    return 0;
}

    就这个题目而言，我写的link-cut-tree的代码要比树链剖分快一些，大概有这么两个原因：①树链剖分或者线段树更新的部分我写得太挫了；②用树链剖分写的话，在求解最大连续和的时候本身就比较麻烦，而link-cut-tree求解的时候则很方便。