树链剖分+线段树 BZOJ 1036 [ZJOI2008]树的统计Count

 

题目链接

题意: I. CHANGE u t : 把结点u的权值改为t

   II. QMAX u v: 询问从点u到点v的路径上的节点的最大权值

   III. QSUM u v: 询问从点u到点v的路径上的节点的权值和 注意:从点u到点v的路径上的节点包括u和v本身

分析:树链剖分第一题,把树拆成一条条链,有重链和轻链,每个点有转换后的新的位置,同一条链是线性的区间,这样可以用线段树进行操作。第一个是单点更新,第二个先求LCA,然后把u和v移动到lca所在的链,转移一次都是转移链,区间询问最大值和总和,第三种类似。

#include <bits/stdc++.h>
 
const int N = 3e4 + 5;
const int D = 20;
const int INF = 0x7fffffff;
struct Edge {
    int v, nex;
}edge[N<<1];
int head[N];
int a[N];
int sz[N], dep[N], belong[N], pos[N];
int rt[N][D];
int n, tote, loc;
 
void add_edge(int u, int v) {
    edge[tote].v = v; edge[tote].nex = head[u];
    head[u] = tote++;
}
 
void init_edge() {
    memset (head, -1, sizeof (head));
    tote = 0;
}
 
void DFS1(int u, int fa) {
    sz[u] = 1; dep[u] = dep[fa] + 1;
    rt[u][0] = fa;
    for (int i=head[u]; ~i; i=edge[i].nex) {
        int v = edge[i].v;
        if (v == fa) {
            continue;
        }
        DFS1 (v, u);
        sz[u] += sz[v];
    }
}
 
void DFS2(int u, int fa, int chain) {
    int k = 0;
    pos[u] = ++loc;
    belong[u] = chain;
    for (int i=head[u]; ~i; i=edge[i].nex) {
        int v = edge[i].v;
        if (v == fa) {
            continue;
        }
        if (sz[v] > sz[k]) {
            k = v;
        }
    }
    if (k == 0) {
        return ;
    }
    DFS2 (k, u, chain);
    for (int i=head[u]; ~i; i=edge[i].nex) {
        int v = edge[i].v;
        if (v == fa || v == k) {
            continue;
        }
        DFS2 (v, u, v);
    }
}
 
void init_LCA() {
    for (int j=1; j<D; ++j) {
        for (int i=1; i<=n; ++i) {
            rt[i][j] = rt[i][j-1] == 0 ? 0 : rt[rt[i][j-1]][j-1];
        }
    }
}
 
int LCA(int u, int v) {
    if (dep[u] < dep[v]) {
        std::swap (u, v);
    }
    for (int i=0; i<D; ++i) {
        if ((dep[u] - dep[v]) >> i & 1) {
            u = rt[u][i];
        }
    }
    if (u == v) {
        return u;
    }
    for (int i=D-1; i>=0; --i) {
        if (rt[u][i] != rt[v][i]) {
            u = rt[u][i];
            v = rt[v][i];
        }
    }
    return rt[u][0];
}
 
int mx[N<<2], sum[N<<2];
 
#define lson l, mid, o << 1
#define rson mid + 1, r, o << 1 | 1
 
void push_up(int o) {
    mx[o] = std::max (mx[o<<1], mx[o<<1|1]);
    sum[o] = sum[o<<1] + sum[o<<1|1];
}
 
void updata(int p, int v, int l, int r, int o) {
    if (l == r) {
        mx[o] = sum[o] = v;
        return ;
    }
    int mid = l + r >> 1;
    if (p <= mid) {
        updata (p, v, lson);
    } else {
        updata (p, v, rson);
    }
    push_up (o);
}
 
int query_max(int ql, int qr, int l, int r, int o) {
    if (ql <= l && r <= qr) {
        return mx[o];
    }
    int mid = l + r >> 1, ret = -INF;
    if (ql <= mid) {
        ret = std::max (ret, query_max (ql, qr, lson));
    }
    if (qr > mid) {
        ret = std::max (ret, query_max (ql, qr, rson));
    }
    return ret;
}
 
int query_sum(int ql, int qr, int l, int r, int o) {
    if (ql <= l && r <= qr) {
        return sum[o];
    }
    int mid = l + r >> 1, ret = 0;
    if (ql <= mid) {
        ret += query_sum (ql, qr, lson);
    }
    if (qr > mid) {
        ret += query_sum (ql, qr, rson);
    }
    return ret;
}
 
int get_sum(int u, int lca) {
    int ret = 0;
    while (belong[u] != belong[lca]) {
        ret += query_sum (pos[belong[u]], pos[u], 1, n, 1);
        u = rt[belong[u]][0];
    }
    ret += query_sum (pos[lca], pos[u], 1, n, 1);
    return ret;
}
 
int get_max(int u, int lca) {
    int ret = -INF;
    while (belong[u] != belong[lca]) {
        ret = std::max (ret, query_max (pos[belong[u]], pos[u], 1, n, 1));
        u = rt[belong[u]][0];
    }
    ret = std::max (ret, query_max (pos[lca], pos[u], 1, n, 1));
    return ret;
}
 
void prepare() {
    sz[0] = 0; dep[0] = 0;
    DFS1 (1, 0);
    loc = 0;
    DFS2 (1, 0, 1);
    init_LCA ();
    for (int i=1; i<=n; ++i) {
        updata (pos[i], a[i], 1, n, 1);
    }
}
 
int main() {
    scanf ("%d", &n);
    init_edge ();
    for (int i=1; i<n; ++i) {
        int u, v;
        scanf ("%d%d", &u, &v);
        add_edge (u, v);
        add_edge (v, u);
    }
    for (int i=1; i<=n; ++i) {
        scanf ("%d", a+i);
    }
 
    prepare ();
 
    int q;
    scanf ("%d", &q);
    char ch[6];
    int x, y;
    while (q--) {
        scanf ("%s%d%d", ch, &x, &y);
        if (ch[0] == 'C') {
            a[x] = y;
            updata (pos[x], a[x], 1, n, 1);
        } else {
            int lca = LCA (x, y);
            if (ch[1] == 'M') {
                printf ("%d\n", std::max (get_max (x, lca), get_max (y, lca)));
            } else {
                printf ("%d\n", get_sum (x, lca) + get_sum (y, lca) - a[lca]);
            }
        }
    }
    return 0;
}

  

posted @ 2016-05-20 13:48  Running_Time  阅读(205)  评论(0编辑  收藏  举报