• hdu 4912 Paths on the tree(树链拆分+贪婪)


    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,而且两两通道不想交。

    解题思路:用树链剖分求LCA,然后依据通道两端节点的LCA深度排序,从深度最大优先选。推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下面点所有标记。

    #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
    ", ans);
        }
        return 0;
    }

    版权声明:本文博客原创文章,博客,未经同意,不得转载。

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  • 原文地址:https://www.cnblogs.com/yxwkf/p/4671060.html
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