Codeforces 191C Fools and Roads(树链拆分)

题目链接：Codeforces 191C Fools and Roads

题目大意：给定一个N节点的数。然后有M次操作，每次从u移动到v。问说每条边被移动过的次数。

解题思路：树链剖分维护边，用一个数组标记就可以，不须要用线段树。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1e5 + 5;
int N, Q, ne, first[maxn], f[maxn], jump[maxn * 2];
int id, idx[maxn], top[maxn], far[maxn], son[maxn], dep[maxn], cnt[maxn];
struct Edge {
    int u, v;
    void set (int u, int v) {
        this->u = u;
        this->v = v;
    }
}ed[maxn * 2];
void dfs (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(v, u, d + 1);
        cnt[u] += cnt[v];
        if (cnt[son[u]] < cnt[v])
            son[u] = v;
    }
}
void dfs(int u, int rot) {
    top[u] = rot;
    idx[u] = ++id;
    if (son[u])
        dfs(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(v, v);
    }
}
inline void add_Edge(int u, int v) {
    ed[ne].set(u, v);
    jump[ne] = first[u];
    first[u] = ne++;
}
void init () {
    int u, v;
    ne = id = 0;
    memset(first, -1, sizeof(first));
    scanf("%d", &N);
    for (int i = 1; i < N; i++) {
        scanf("%d%d", &u, &v);
        add_Edge(u, v);
        add_Edge(v, u);
    }
    dfs(1, 0, 0);
    dfs(1, 1);
    for (int i = 0; i < N - 1; i++) {
        int t = i * 2;
        if (dep[ed[t].u] < dep[ed[t].v])
            swap(ed[t].u, ed[t].v);
    }
}
inline void add (int l, int r) {
    f[l]++, f[r + 1]--;
}
void solve (int u, int v) {
    int p = top[u], q = top[v];
    while (p != q) {
        if (dep[p] < dep[q]) {
            swap(p, q);
            swap(u, v);
        }
        add(idx[p], idx[u]);
        u = far[p];
        p = top[u];
    }
    if (u == v)
        return;
    if (dep[u] > dep[v])
        swap(u, v);
    add(idx[son[u]], idx[v]);
}
int main () {
    init();
    scanf("%d", &Q);
    int u, v;
    while (Q--) {
        scanf("%d%d", &u, &v);
        solve(u, v);
    }
    int mv = 0;
    for (int i = 1; i <= N; i++) {
        mv += f[i];
        f[i] = mv;
    }
    printf("%d", f[idx[ed[0].u]]);
    for (int i = 1; i < N - 1; i++)
        printf(" %d", f[idx[ed[i*2].u]]);
    printf("\n");
    return 0;
}

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