hdu 5242(贪心)

来源：互联网发布：图片纠偏软件编辑：程序博客网时间：2024/06/14 21:29

题意：给一颗N个节点的树，有N-1条有向边，每个点有一个权值，从1号点放k个人走到叶子结点，求覆盖路径的最大权值和。

思路：先反向建树，求出每个点到根节点的权值和，可以保证这是一条无分岔的路。然后从大到小排序，从权值和大的点开始覆盖道路，最后再排序取前k大的路即可。

#include<bits/stdc++.h>using namespace std;typedef long long ll;typedef pair<ll,int> P;#define fi first#define se second#define INF 0x3f3f3f3f#define clr(x,y) memset(x,y,sizeof x)#define PI acos(-1.0)#define ITER set<int>::iteratorconst int Mod = 1e9 + 7;const int maxn = 1e5 + 10;ll a[maxn];struct Edge{int to,next;}edge[maxn << 2];int head[maxn],edge_num;void Init(){    clr(head,-1);edge_num = 0;}void add_edge(int x,int y){edge[edge_num] = (Edge){y,head[x]};head[x] = edge_num ++;}int id[maxn];ll val[maxn];ll dfs(int u){    if(val[u] != -1)return val[u];    ll ret = a[u];    for(int i = head[u];i != -1;i = edge[i].next)    {        int v = edge[i].to;ret += dfs(v);    }    return val[u] = ret;}bool cmp(int x,int y){return val[x] > val[y];}bool vis[maxn];ll dfs2(int u){    if(vis[u])return 0;vis[u] = true;    ll ret = a[u];    for(int i = head[u];i != -1;i = edge[i].next)    {        int v = edge[i].to;        ret += dfs2(v);    }    return ret;}int main(){    int Tcase;scanf("%d",&Tcase);    for(int ii = 1;ii <= Tcase;ii ++)    {        Init();        int n,k;scanf("%d%d",&n,&k);        for(int i = 1; i <= n;i ++)scanf("%lld",&a[i]);        for(int i = 1;i <= n - 1;i ++){int x,y;scanf("%d%d",&x,&y);add_edge(y,x);}        clr(val,-1);        for(int i = 1;i <= n; i ++)        {            id[i] = i;val[i] = dfs(i);        }        sort(id + 1,id + n  + 1,cmp);//        for(int i = 1;i <= n;i ++)cout << id[i] <<" -> " << val[id[i]] << endl;        clr(vis,false);        for(int i = 1;i <= n;i ++)        {            int pos = id[i];            val[pos] = dfs2(pos);        }        ll ans = 0;        sort(val + 1,val + n + 1,greater<ll>());        for(int i = 1;i <= k;i ++)ans += val[i];        printf("Case #%d: %lld\n",ii,ans);    }    return 0;}

阅读全文

0 0