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);

        }



    }

}

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