HDU 4169 Wealthy Family 优先队列

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题目大意：

从一棵树的N <= 150000个结点中选出K个使得权值和最大, 选出的K个点两两之间都不是祖先关系

大致思路：

HDU 4233的简化版....

HDU4233题解见这里

代码如下：

Result : Accepted Memory : 8444 KB Time : 624 ms

/* * Author: Gatevin * Created Time:  2015/8/10 20:10:18 * File Name: Sakura_Chiyo.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;#define maxn 150010int N, K, root;int w[maxn];bool select[maxn];vector<int> G[maxn];int f[maxn];//f[u]表示从u的子孙节点中选两个权值和的最大值void dfs(int now)//处理出每个子树u及其子树中最小的权值w[u], 以及其子树中两个没有关联的父亲关系的点权值最大和{    int nex;    int mx = -1e9;//表示前几个子树中的最小w    for(int i = 0, sz = G[now].size(); i < sz; i++)    {        nex = G[now][i];        dfs(G[now][i]);        f[now] = max(f[now], max(f[nex], w[nex] + mx));        w[now] = max(w[now], w[nex]);        mx = max(mx, w[nex]);    }    return;}void solve(){    memset(select, 0, sizeof(select));    priority_queue<pair<int, int> > Q;    Q.push(make_pair(w[root], root));    int ret = 0;    while(K)    {        if(Q.empty())//选不出K个        {            puts("impossible");            return;        }        int u = Q.top().second;        Q.pop();        if(select[u])//说明现在选择u的子节点中的两个要更优        {            K++;            select[u] = 0;//放弃选择子树的根节点u            ret -= w[u];            for(int i = 0, sz = G[u].size(); i < sz; i++) Q.push(make_pair(w[G[u][i]], G[u][i]));//加入其儿子节点        }        else        {            K--;            select[u] = 1;            ret += w[u];//选择了u结点            Q.push(make_pair(f[u] - w[u], u));//那么当选择两个其子节点和和它的差值要加入队列, 以确定是否要放弃这个结点        }    }    printf("%d\n", ret);    return;}int main(){    while(~scanf("%d %d", &N, &K))    {        for(int i = 1; i <= N; i++) G[i].clear();        int p;        for(int i = 1; i <= N; i++)        {            f[i] = -1e9;            scanf("%d %d", &p, w + i);            if(p == 0) root = i;            else G[p].push_back(i);        }        dfs(root);        solve();    }    return 0;}

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