hdu 4912 Paths on the tree(树链剖分+贪心)

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题目链接：hdu 4912 Paths on the tree

题目大意：给定一棵树，和若干个通道，要求尽量选出多的通道，并且两两通道不想交。

解题思路：用树链剖分求LCA，然后根据通道两端节点的LCA深度排序，从深度最大优先选，判断两个节点均没被标

记即为可选通道。每次选完通道，将该通道LCA以下点全部标记。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 1e5 + 5;

int N, M, E, first[maxn], jump[maxn * 2], link[maxn * 2], vis[maxn];
int id, idx[maxn], top[maxn], far[maxn], son[maxn], cnt[maxn], dep[maxn];

inline void add_Edge (int u, int v) {
    link[E] = v;
    jump[E] = first[u];
    first[u] = E++;
}

void dfs (int u, int pre, int d) {
    far[u] = pre;
    son[u] = 0;
    cnt[u] = 1;
    dep[u] = d;
    for (int i = first[u]; i + 1; i = jump[i]) {
        int v = link[i];
        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 = link[i];
        if (v == far[u] || v == son[u])
            continue;
        dfs(v, v);
    }
}

void dfs (int u) {
    if (vis[u])
        return;
    vis[u] = 1;
    for (int i = first[u]; i + 1; i = jump[i]) {
        int v = link[i];
        if (v == far[u])
            continue;
        dfs(v);
    }
}

struct query {
    int u, v, r, d;
    void set(int u, int v, int r, int d) {
        this->u = u;
        this->v = v;
        this->r = r;
        this->d = d;
    }
    friend bool operator < (const query& a, const query& b) {
        return a.d > b.d;
    }
}q[maxn];

int LCA (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);
        }
        u = far[p];
        p = top[u];
    }
    return dep[u] < dep[v] ? u : v;
}

void init () {
    E = id = 0;
    memset(first, -1, sizeof(first));

    int u, v;
    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 < M; i++) {
        scanf("%d%d", &u, &v);
        int rot = LCA(u, v);
        q[i].set(u, v, rot, dep[rot]);
    }
    sort(q, q + M);
}

int main () {
    while (scanf("%d%d", &N, &M) == 2) {
        init();

        int ans = 0;
        memset(vis, 0, sizeof(vis));
        for (int i = 0; i < M; i++) {
            if (vis[q[i].u] || vis[q[i].v])
                continue;
            dfs(q[i].r);
            ans++;
        }
        printf("%d\n", ans);
    }
    return 0;
}

hdu 4912 Paths on the tree(树链剖分+贪心)

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原文地址：http://blog.csdn.net/keshuai19940722/article/details/40189833