HDU - 3966 Aragorn's Story(树剖模版题)

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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, 1void 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1,1void 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;}

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