codeforces 165D.Beard Graph 解题报告

题意：

     给一棵树，树的每条边有一种颜色，黑色或白色，一开始所有边均为黑色，有两个操作：

     操作1：将第i条边变成白色或将第i条边变成黑色。

     操作2 ：询问u，v两点之间仅经过黑色变的最短距离。

 

树链剖分+树状数组

学习树链剖分：

/*
       树链剖分：
       划分轻重链，效果是将一颗树变成了若干段连续的区间。
       向上记录边权
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

const int MAX = 111111;
//子树结点数,所在链链头,同链儿子,结点深度,节点在区间的位置,父节点
int siz[MAX], top[MAX], son[MAX], dep[MAX], w[MAX], fa[MAX];
//链式前向星记录一颗树
struct Edge {
    int v, ne, id;
} edge[MAX << 1];

int head[MAX], cnt, num;
int vis[MAX], pos[MAX];;
//第一次dfs,提取基本信息,划分轻重链
void Dfs1 (int u, int v) {
    fa[v] = u, dep[v] = dep[u] + 1;
    siz[v] = 1;
    vis[v] = 1;
    int tem = 0, p = -1;
    for (int i = head[v]; i != 0; i = edge[i].ne) {
        int kid = edge[i].v;
        if (!vis[kid]) {
            Dfs1 (v, kid);
            siz[u] += siz[kid];
            if (tem < siz[kid]) tem = siz[kid], p = kid;
        }
    }
    son[v] = p;
}
//第二次DFS,将重链映射到区间
void Dfs2 (int h, int v) {
    top[v] = h;
    w[v] = ++num;
    vis[v] = 1;
    if (son[v] != -1) Dfs2 (h, son[v]);
    for (int i = head[v]; i != 0; i = edge[i].ne) {
        int kid = edge[i].v;
        if (son[v] != kid && !vis[kid])
            Dfs2 (kid, kid);
              pos[edge[i].id] = w[kid]; //边映射到下结点
    }
}

void addE (int u, int v, int num) {
    edge[++cnt].v = v, edge[cnt].id = num;
    edge[cnt].ne = head[u], head[u] = cnt;
}

int A[MAX]; //pos为边的新编号
void add (int x, int k) {
    for (; x <= num; x += x & -x) A[x] += k;
}
int getsum (int x) {
    int s = 0;
    for (; x > 0; x -= x & -x)  s += A[x];
    return s;
}
void modify (int x, int k) {
    x = pos[x];
    int sta = getsum (x) - getsum (x - 1);
    if (sta == k) return;
    else add (x, k - sta);
}

void query (int u, int v) {
    int s = 0, f1 = top[u], f2 = top[v], len = 0;
    while (f1 != f2) {
        if (dep[f1] < dep[f2]) {
            int y = w[v], x = w[f2];
            if (f2 != v) {
                if (s += getsum (y) - getsum (x) ) break;
                v = f2;
                len += y - x;
            }
            if (s += getsum (x) - getsum (x - 1) ) break;
            v = fa[v], f2 = top[v];
            len++;
        }
        else {
            int y = w[u], x = w[f1];
            if (f1 != u) {
                if (s += getsum (y) - getsum (x) ) break;
                u = f1;
                len += y - x;
            }
            if (s += getsum (x) - getsum (x - 1) ) break;
            u = fa[u], f1 = top[u];
            len++;
        }
    }
    int y = w[v], x = w[u];
    s += getsum (max (y, x) ) - getsum (min (y, x) );
    len += abs (y - x);
    printf ("%d\n", s == 0 ? len : -1);
}
int n, u, v, m, t;
int main() {
    scanf ("%d", &n);
    for (int i = 1; i < n; i++) {
        scanf ("%d %d", &u, &v);
        addE (u, v, i), addE (v, u, i);
    }
    Dfs1 (0, 1);
    memset (vis, 0, sizeof vis);
    Dfs2 (1, 1);
    scanf ("%d", &m);
    int k, l, r,tol=0;
    for (int i = 1; i <= m; i++) {
        scanf ("%d", &k);
        if (k != 3) {
            scanf ("%d", &t);
            modify (t, k == 1 ? 0 : 1);
        }
        if (k == 3) {
            scanf ("%d %d", &l, &r);
                     query (l, r);
        }

    }
}