HDU5242：Game（树上贪心）

Game

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1465    Accepted Submission(s): 477

Problem Description

It is well known that Keima Katsuragi is The Capturing God because of his exceptional skills and experience in ''capturing'' virtual girls in gal games. He is able to play k games simultaneously.

One day he gets a new gal game named ''XX island''. There are n scenes in that game, and one scene will be transformed to different scenes by choosing different options while playing the game. All the scenes form a structure like a rooted tree such that the root is exactly the opening scene while leaves are all the ending scenes. Each scene has a value , and we use wi as the value of the i-th scene. Once Katsuragi entering some new scene, he will get the value of that scene. However, even if Katsuragi enters some scenes for more than once, he will get wi for only once.

For his outstanding ability in playing gal games, Katsuragi is able to play the game k times simultaneously. Now you are asked to calculate the maximum total value he will get by playing that game for k times.

 

Input

The first line contains an integer T(T≤20), denoting the number of test cases.

For each test case, the first line contains two numbers n,k(1≤k≤n≤100000), denoting the total number of scenes and the maximum times for Katsuragi to play the game ''XX island''.

The second line contains n non-negative numbers, separated by space. The i-th number denotes the value of the i-th scene. It is guaranteed that all the values are less than or equal to 231−1.

In the following n−1 lines, each line contains two integers a,b(1≤a,b≤n), implying we can transform from the a-th scene to the b-th scene.

We assume the first scene(i.e., the scene with index one) to be the opening scene(i.e., the root of the tree).

 

Output

For each test case, output ''Case #t:'' to represent the t-th case, and then output the maximum total value Katsuragi will get.

 

Sample Input

25 24 3 2 1 11 21 52 32 45 34 3 2 1 11 21 52 32 4

 

Sample Output

Case #1: 10Case #2: 11

 

Source

The 2015 ACM-ICPC China Shanghai Metropolitan Programming Contest

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 <iostream># include <cstdio># include <cstring># include <algorithm># define ll long longusing namespace std;const int maxn = 1e5;int cnt, cas=1;ll val[maxn+3], ne[maxn+3], vis[maxn+3], road[maxn+3], id[maxn+3];struct node{    int to, next;}a[maxn+3];void add_edge(int u, int v){    a[cnt].to = v;    a[cnt].next = ne[u];    ne[u] = cnt++;}ll dfs(int u)//所有点到根节点的权值和。{    if(vis[u]) return road[u];    vis[u] = 1;    road[u] = val[u];    for(int i=ne[u]; i!=-1; i=a[i].next)    {        int v = a[i].to;        road[u] += dfs(v);    }    return road[u];}ll dfs2(int u){    if(vis[u]) return 0;    vis[u] = 1;    ll tmp = val[u];    for(int i=ne[u]; i!=-1; i=a[i].next)    {        int v = a[i].to;        tmp += dfs2(v);    }    return tmp;}bool cmp(int x, int y){    return road[x] > road[y];}int main(){    int t, n, k, a, b;    scanf("%d",&t);    while(t--)    {        cnt = 0;        memset(ne, -1, sizeof(ne));        scanf("%d%d",&n,&k);        for(int i=1; i<=n; ++i)            scanf("%lld",&val[i]);        for(int i=1; i<n; ++i)        {            scanf("%d%d",&a,&b);            add_edge(b, a);        }        memset(vis, 0, sizeof(vis));        memset(road, 0, sizeof(road));        for(int i=1; i<=n; ++i)        {            id[i] = i;            road[i] = dfs(i);        }        sort(id+1, id+1+n, cmp);        memset(vis, 0, sizeof(vis));        memset(road, 0, sizeof(road));        for(int i=1; i<=n; ++i)        {            int now = id[i];            road[i] = dfs2(now);        }        sort(road+1, road+1+n, greater<ll>());        ll ans = 0;        for(int i=1; i<=k; ++i)            ans += road[i];        printf("Case #%d: %lld\n",cas++, ans);    }    return 0;}