多校第三场 1005 HDU 6060 思维贪心+dfs+一维建树

RXD and dividing

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1353    Accepted Submission(s): 244

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 ⋃Si={2,3,…,n} and for all different i,j , we can conclude that Si⋂Sj=∅. 
Then he calulates res=∑ki=1f({1}⋃Si).
He wants to maximize the res.
1≤k≤n≤106
the cost of each edge∈[1,105]
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 41 2 32 3 42 4 52 5 6

 

Sample Output

27

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<vector>#include<map>#include<string>#include<cmath>#include<vector>#define maxlen 1000000+7using namespace std;vector<pair<int, int> >nextpoint[maxlen];bool vis[maxlen];/*考虑每条边对答案的贡献，很容易可以知道能用到这条边的最多的生成树是从这条边往下的点的数量，即每个下面的点都是独立的点集的时候。但是由于最多只有k个集合 所以答案就是w[i]*min(k,sz[i])对i求和建树用vector+pair建统计下面节点数量用dfs + vis数组注意结果中间运算用long long维护*/int n, k;long long ans;int dfs(int x){int cnt = 1;for(int i = 0; i < nextpoint[x].size(); i++){int newpos = nextpoint[x][i].first;if(vis[newpos])continue;vis[newpos] = 1;int amount = dfs(newpos);//cout<<amount<<endl;ans += (long long)min(amount, k) * (long long)nextpoint[x][i].second;cnt += amount;}return cnt;}int main(){//freopen("test.txt","r",stdin);while(scanf("%d %d", &n, &k) != EOF){for(int i = 1; i <= n; i++)nextpoint[i].clear();memset(vis, 0, sizeof vis);ans = 0;for(int i = 0; i < n - 1; i++){int x;int y;int val;scanf("%d %d %d", &x, &y, &val);nextpoint[x].push_back(pair<int, int>(y, val));nextpoint[y].push_back(pair<int, int>(x, val));}vis[1]=1;dfs(1);printf("%lld\n", ans);}return 0;}