【思维】hdu 6060 RXD and dividing

数组模拟链表：http://blog.csdn.net/baidu_35643793/article/details/76850687

RXD and dividing

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1660    Accepted Submission(s): 713

Problem Description

RXD has a tree T, with the size of n. Each edge has a cost.
Define f(S) as the the cost of the minimal Steiner Tree of the set S on tree T. 
he wants to divide 2,3,4,5,6,…n into k parts S1,S2,S3,…Sk,
where    and for all different i,j , we can conclude that 
Then he calulates
He wants to maximize the res.
1≤k≤n≤106
the cost of each edge∈[1,1e5]
Si might be empty.
f(S) means that you need to choose a couple of edges on the tree to make all the points in S connected, and you need to minimize the sum of the cost of these edges. f(S) is equal to the minimal cost 
Input
There are several test cases, please keep reading until EOF.
For each test case, the first line consists of 2 integer n,k, which means the number of the tree nodes , and k means the number of parts.
The next n−1 lines consists of 2 integers, a,b,c, means a tree edge (a,b) with cost c.
It is guaranteed that the edges would form a tree.
There are 4 big test cases and 50 small test cases.
small test case means n≤100.
Output
For each test case, output an integer, which means the answer.
Sample Input
5 4
1 2 3
2 3 4
2 4 5
2 5 6
Sample Output
27

题意：有一个n各节点的树，现将节点2-n分为k组，定义每组的权值为该组内所有点加编号为1的节点相互连接所经过的边的权值之和，求k组点集最大的和；

思路：思考方向：每条边的贡献，（发现这个关于贡献的思考方向很重要 ）

题解思路；

dif ( i ) : i 节点子树内不同的标号数目；

sz( i ) : i 节点子树的大小；

(x,fax) : 节点x和它的父亲所组成的边；

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;const int N=2001000;int head[N+10];struct node{    int v,w;    int next;} edge[N+10];int n,m,cnt=0;ll ans=0;void add_edge(int from,int to,int val)  //创建邻接表{    edge[++cnt].v=to;    edge[cnt].w=val;    edge[cnt].next=head[from];    head[from]=cnt;}int dfs(int fa,int x,int val)  //dfs求各子树的大小；{    int sum=1;    for(int i=head[x];i;i=edge[i].next)    {        int to=edge[i].v;        int val=edge[i].w;        if(to!=fa)            sum+=dfs(x,to,val);    }    if(sum>m)        ans+=(ll)m*val;    else        ans+=(ll)sum*val;    return sum;}int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {        cnt=0;        memset(head,0,sizeof(head));        for(int i=1; i<=n-1; i++)        {            int u,v,w;            scanf("%d%d%d",&u,&v,&w);            add_edge(u,v,w);            add_edge(v,u,w);        }        ans=0;        dfs(0,1,0);  //        printf("%I64d\n",ans);    }    return 0;}

参考：http://blog.csdn.net/your_eyes_see_sea/article/details/76578077、http://blog.csdn.net/kkkkahlua/article/details/76562559