Codeforces 191c fools and roads)

Link: codeforces 191c fools and roads

Given the number of N nodes, there are m operations. Each time you move from u to V, the number of times each edge is moved.

Solution: Tree links are divided into maintenance edges and can be marked with a number group. Line Segment trees are not required.

#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;}

Codeforces 191c fools and roads)