HDU 4912 （贪心）

最新推荐文章于 2020-04-20 14:05:09 发布

原创最新推荐文章于 2020-04-20 14:05:09 发布 · 454 阅读

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贪心同时被 2 个专栏收录

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LCA

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本文介绍了一种基于树形结构的贪心算法问题解决方法，通过LCA算法找到链的最大不相交集合。

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题目链接：点击这里

题意：在一棵树上给出m个链，求出最多能够选择多少个链使得所有的链都没有交点。

根据lca自底向上贪心即可。

#include <cstring>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;
#define maxn 100005
#define maxm maxn*2

int n, m;
struct node {
    int v, next;
}edge[maxm];
int cnt, head[maxn];
int deep[maxn];
struct Q {
    int u, v;
    int lca;
    int next;
    bool operator < (const Q &a) const {
        return deep[lca] > deep[a.lca];
    }
}qu[maxm];
int cnt2, head2[maxn];
int p[maxn], fa[maxn];
bool vis[maxn];

#define find Find
int find (int x) {
    return p[x] == x ? p[x] : p[x] = find (p[x]);
}

void add_edge (int u, int v) {
    edge[cnt].v = v, edge[cnt].next = head[u], head[u] = cnt++;
}

void add_query (int u, int v) {
    qu[cnt2].u = u, qu[cnt2].v = v, qu[cnt2].next = head2[u], head2[u] = cnt2++;
}

void LCA (int u, int father, int d) {
    deep[u] = d;
    p[u] = u;
    fa[u] = father;
    vis[u] = 1;
    for (int i = head[u]; i != -1; i = edge[i].next) {
        int v = edge[i].v;
        if (vis[v]) continue;
        LCA (v, u, d+1);
        p[v] = u;
    }
    for (int i = head2[u]; i != -1; i = qu[i].next) {
        int v = qu[i].v;
        if (vis[v]) {
            qu[i^1].lca = qu[i].lca = find (v);
        }
    }
}

void del (int u) {
    vis[u] = 1;
    for (int i = head[u]; i != -1; i = edge[i].next) {
        int v = edge[i].v;
        if (vis[v] || v == fa[u]) continue;
        del (v);
    }
}

void solve () {
    int ans = 0;
    for (int i = 0; i < cnt2; i++) {
        if (vis[qu[i].u] || vis[qu[i].v]) continue;
        ans++;
        del (qu[i].lca);
    }
    printf ("%d\n", ans);
}

int main () {
    while (scanf ("%d%d", &n, &m) == 2) {
        cnt = cnt2 = 0;
        for (int i = 1; i <= n; i++) head[i] = head2[i] = -1, vis[i] = 0;
        for (int i = 1; i < n; i++) {
            int u, v;
            scanf ("%d%d", &u, &v);
            add_edge (u, v);
            add_edge (v, u);
        }
        for (int i = 1; i <= m; i++) {
            int u, v;
            scanf ("%d%d", &u, &v);
            add_query (u, v);
            add_query (v, u);
        }
        LCA (1, 1, 0);
        cnt2 >>= 1;
        for (int i = 0; i < cnt2; i++) {
            qu[i] = qu[i<<1];
        }
        sort (qu, qu+cnt2);
        memset (vis, 0, sizeof vis);
        solve ();
    }
    return 0;
}