[洛谷P2590][ZJOI2008]树的统计

最新推荐文章于 2021-02-21 12:35:22 发布

转载最新推荐文章于 2021-02-21 12:35:22 发布 · 138 阅读

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原文链接：http://www.cnblogs.com/Memory-of-winter/p/9483053.html

文章标签：

#c/c++

本文详细解析了一道涉及树链剖分和线段树的经典算法题目，介绍了如何利用树链剖分和线段树实现节点权值的快速修改及路径上节点最大值与权值和的查询。

题目大意：一棵树，支持三个操作，

$CHANGE\;u\;t:$ 把结点$u$的权值改为$t$

$QMAX\;u\;v:$ 询问从点$u$到点$v$的路径上的节点的最大权值

$QSUM\;u\;v:$ 询问从点$u$到点$v$的路径上的节点的权值和

题解：裸的树链剖分

卡点：线段树区间修改我不知道哪根筋搭错了，写了$l\;==\;r$（应为$L\;\leq\;l\;\&\&\;R\;\geq\;r$）

C++ Code：

#include <cstdio>
#define maxn 30010
using namespace std;
const int inf = 0x3f3f3f3f;
int n, m;
int w[maxn], V[maxn << 2], M[maxn << 2];
inline int max(int a, int b) {return a > b ? a : b;}
void swap(int &a, int &b) {a ^= b ^= a ^= b;}
void update(int rt) {
    V[rt] = V[rt << 1] + V[rt << 1 | 1];
    M[rt] = max(M[rt << 1], M[rt << 1 | 1]);
}
void build(int rt, int l, int r) {
    if (l == r) {
        V[rt] = M[rt] = w[l];
        return ;
    }
    int mid = l + r >> 1;
    build(rt << 1, l, mid);
    build(rt << 1 | 1, mid + 1, r);
    update(rt);
}
void add(int rt, int l, int r, int p, int num) {
    if (l == r) {
        V[rt] = M[rt] = num;
        return ;
    }
    int mid = l + r >> 1;
    if (p <= mid) add(rt << 1, l, mid, p, num);
    else add(rt << 1 | 1, mid + 1, r, p, num);
    update(rt);
}
int askS(int rt, int l, int r, int L, int R) {
    if (L <= l && R >= r) {
        return V[rt];
    }
    int mid = l + r >> 1, ans = 0;
    if (L <= mid) ans = askS(rt << 1, l, mid, L, R);
    if (R > mid) ans = ans + askS(rt << 1 | 1, mid + 1, r, L, R);
    return ans;
}
int askM(int rt, int l, int r, int L, int R) {
    if (L <= l && R >= r) {
        return M[rt];
    }
    int mid = l + r >> 1, ans = -inf;
    if (L <= mid) ans = askM(rt << 1, l, mid, L, R);
    if (R > mid) ans = max(ans, askM(rt << 1 | 1, mid + 1, r, L, R));
    return ans;
}

int head[maxn], cnt;
struct Edge {
    int to, nxt;
} e[maxn << 1];
void addE(int a, int b) {
    e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
}
int fa[maxn], sz[maxn], son[maxn], dep[maxn];
int top[maxn], dfn[maxn], idx;
void dfs1(int rt) {
    sz[rt] = 1;
    for (int i = head[rt]; i; i = e[i].nxt) {
        int v = e[i].to;
        if (v != fa[rt]) {
            dep[v] = dep[rt] + 1;
            fa[v] = rt;
            dfs1(v);
            if (!son[rt] || sz[v] > sz[son[rt]]) son[rt] = v;
            sz[rt] += sz[v];
        }
    }
}
void dfs2(int rt) {
    dfn[rt] = ++idx;
    int v = son[rt];
    if (v) top[v] = top[rt], dfs2(v);
    for (int i = head[rt]; i; i = e[i].nxt) {
        v = e[i].to;
        if (v != son[rt] && v != fa[rt]) {
            top[v] = v;
            dfs2(v);
        }
    }
}
int queryS(int x, int y) {
    int ans = 0;
    while (top[x] != top[y]) {
        if (dep[top[x]] < dep[top[y]]) swap(x, y);
        ans += askS(1, 1, n, dfn[top[x]], dfn[x]);
        x = fa[top[x]];
    }
    if (dep[x] > dep[y]) swap(x, y);
    ans += askS(1, 1, n, dfn[x], dfn[y]);
    return ans;
}
int queryM(int x, int y) {
    int ans = -inf;
    while (top[x] != top[y]) {
        if (dep[top[x]] < dep[top[y]]) swap(x, y);
        ans = max(ans, askM(1, 1, n, dfn[top[x]], dfn[x]));
        x = fa[top[x]];
    }
    if (dep[x] > dep[y]) swap(x, y);
    ans = max(ans, askM(1, 1, n, dfn[x], dfn[y]));
    return ans;
}
int main() {
    scanf("%d", &n);
    for (int i = 1; i < n; i++) {
        int a, b;
        scanf("%d%d", &a, &b);
        addE(a, b);
        addE(b, a);
    }
    dep[1] = 1;
    dfs1(1);
    top[1] = 1;
    dfs2(1);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &w[dfn[i]]);
    }
    build(1, 1, n);
    scanf("%d", &m);
    while (m --> 0) {
        char op[10];
        int x, y;
        scanf("%s%d%d", op, &x, &y);
        if (op[1] == 'M') {
            printf("%d\n", queryM(x, y));
        }
        if (op[1] == 'S') {
            printf("%d\n", queryS(x, y));
        }
        if (op[1] == 'H') {
            add(1, 1, n, dfn[x], y);
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/Memory-of-winter/p/9483053.html