HDU - 3966 Aragorn's Story(树剖模版题)

题目大意：给出N个点，有三种操作
1. I a b，第a个营地增加b个人
2. D a b，第a个营地减少b个人
3. Q a，第a个营地有多少人

解题思路：树剖模版题

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int N = 50005;
inline int lowbit(int x) {return x & (-x); }

int dep[N], f[N], son[N], size[N], top[N], id[N], bit[N], val[N], idx;
int n, m, p;
vector<int> g[N];

//dep存储的是节点的深度
//size记录的是以u为根结点的子树的节点数
//f数组记录的是每个节点的父节点
//son记录的是与u同在一重链上的u的子节点
//初始值1, -1, 1

void dfs1(int u, int fa, int depth) {
    dep[u] = depth;
    size[u] = 1;
    f[u] = fa;
    son[u] = 0;
    for (int i = 0; i < g[u].size(); i++) {
        int v = g[u][i];
        if (v == fa) continue;
        dfs1(v, u, depth + 1);
        size[u] += size[v];
        if (size[son[u]] < size[v]) son[u] = v;
    }
}

//top记录的是结点所在的链的顶端结点
//id记录的是树的结点对应的的值
//初始值是1,1

void dfs2(int u, int tp) {
    id[u] = ++idx;
    top[u] = tp;
    //son[u]是重链
    if (son[u]) dfs2(son[u], tp);
    for (int i = 0; i < g[u].size(); i++) {
        int v = g[u][i];
        if (v == f[u] || v == son[u]) continue;
        dfs2(v, v);
    }
}

void add(int x, int v) {
    while (x < N) {
        bit[x] += v;
        x += lowbit(x);
    }
}

void add(int l, int r, int v) {
    add(l, v);
    add(r + 1, -v);
}

void gao(int u, int v, int w) {
    int tp1 = top[u], tp2 = top[v];
    while (tp1 != tp2) {
        if (dep[tp1] < dep[tp2]) {
            swap(tp1, tp2);
            swap(u, v);
        }
        add(id[tp1], id[u], w);
        u = f[tp1];
        tp1 = top[u];
    }
    if (dep[u] > dep[v]) swap(u, v);
    add(id[u], id[v], w);
}

int query(int x) {
    int ans = 0;
    while (x) {
        ans += bit[x];
        x -= lowbit(x);
    }
    return ans;
}

int main() {
    while (scanf("%d%d%d", &n, &m, &p) != EOF) {
        idx = 0;
        memset(bit, 0, sizeof(bit));
        for (int i = 1; i <= n; i++) {
            scanf("%d", &val[i]);
            g[i].clear();
        }

        int u, v;
        while (m--) {
            scanf("%d%d", &u, &v);
            g[u].push_back(v);
            g[v].push_back(u);
        }
        dfs1(1, -1, 1);
        dfs2(1, 1);
        for (int i = 1; i <= n; i++) add(id[i], id[i], val[i]);
        char q[2];
        int a, b, c;
        while (p--) {
            scanf("%s", q);
            if (q[0] == 'I' || q[0] == 'D') {
                scanf("%d%d%d", &a, &b, &c);
                if (q[0] == 'D') c = -c;
                gao(a, b, c);
            }
            else {
                scanf("%d", &a);
                printf("%d\n", query(id[a]));
            }
        }
    }
    return 0;
}