题目链接：poj 2763 Housewife Wind

题目大意：给定一棵树，然后2种操作：

• 0 u：输出路径s到u的权值和，并且s变成u
• 1 i w：节点i增加w

解题思路：树链剖分，然后用线段树维护，单点修改区间查询。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 100005;
int N, Q, S, ne;

int first[maxn], jump[maxn * 2];
int id, far[maxn], son[maxn], cnt[maxn], dep[maxn], top[maxn], idx[maxn];

struct Edge {
    int u, v, w;
    void set (int u, int v, int w) {
        this->u = u;
        this->v = v;
        this->w = w;
    }
}ed[maxn * 2];

void dfs_fir (int u, int pre, int d) {
    far[u] = pre;
    dep[u] = d;
    cnt[u] = 1;
    son[u] = 0;

    for (int i = first[u]; i + 1; i = jump[i]) {
        int v = ed[i].v;
        if (v == pre)
            continue;
        dfs_fir(v, u, d+1);
        cnt[u] += cnt[v];
        if (cnt[son[u]] < cnt[v])
            son[u] = v;
    }
}

void dfs_sec(int u, int rot) {
    top[u] = rot;
    idx[u] = id++;

    if (son[u])
        dfs_sec(son[u], rot);

    for (int i = first[u]; i + 1; i = jump[i]) {
        int v = ed[i].v;
        if (v == far[u] || v == son[u])
            continue;
        dfs_sec(v, v);
    }
}

inline void add_Edge(int u, int v, int w) {
    ed[ne].set(u, v, w);
    jump[ne] = first[u];
    first[u] = ne++;
}

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
typedef long long ll;
int lc[maxn << 2], rc[maxn << 2];
ll s[maxn << 2];

inline void pushup(int u) {
    s[u] = s[lson(u)] + s[rson(u)];
}

void build (int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;
    s[u] = 0;

    if (l == r)
        return;
    int mid = (l + r) / 2;
    build(lson(u), l, mid);
    build(rson(u), mid + 1, r);
    pushup(u);
}

void modify (int u, int x, int w) {
    if (lc[u] == x && rc[u] == x) {
        s[u] = w;
        return;
    }

    int mid = (lc[u] + rc[u]) / 2;
    if (x <= mid)
        modify(lson(u), x, w);
    else
        modify(rson(u), x, w);
    pushup(u);
}

ll query(int u, int l, int r) {
    if (l <= lc[u] && rc[u] <= r)
        return s[u];

    ll ret = 0;
    int mid = (lc[u] + rc[u]) / 2;
    if (l <= mid)
        ret += query(lson(u), l, r);
    if (r > mid)
        ret += query(rson(u), l, r);
    return ret;
}

void init () {
    ne = 0;
    id = 1;

    int u, v, w;
    memset(first, -1, sizeof(first));
    for (int i = 0; i < N - 1; i++) {
        scanf("%d%d%d", &u, &v, &w);
        add_Edge(u, v, w);
        add_Edge(v, u, w);
    }
    dfs_fir(1, 0, 0);
    dfs_sec(1, 1);

    build(1, 1, N);
    for (int i = 0; i < N - 1; i++) {
        if (dep[ed[i*2].u] < dep[ed[i*2].v])
            swap(ed[i*2].u, ed[i*2].v);
        modify(1, idx[ed[i*2].u], ed[i*2].w);
    }
}

ll solve (int u, int v) {
    ll ret = 0;
    int p = top[u], q = top[v];
    while (p != q) {
        if (dep[p] < dep[q]) {
            swap(p, q);
            swap(u, v);
        }
        ret += query(1, idx[p], idx[u]);
        u = far[p];
        p = top[u];
    }

    if (u == v)
        return ret;

    if (dep[u] > dep[v])
        swap(u, v);
    ret += query(1, idx[son[u]], idx[v]);
    return ret;
}

int main () {
    while (scanf("%d%d%d", &N, &Q, &S) == 3) {
        init();
        int k, u, w;
        while (Q--) {
            scanf("%d%d", &k, &u);
            if (k) {
                scanf("%d", &w);
                modify(1, idx[ed[u*2 - 2].u], w);
            } else {
                printf("%lld\n", solve(S, u));
                S = u;
            }
        }
    }
    return 0;
}

poj 2763 Housewife Wind(树链剖分)