2017 Multi-University Training Contest - Team 3 hdu6060 RXD and dividing
地址:http://acm.split.hdu.edu.cn/showproblem.php?pid=6060
题目:
RXD and dividing
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 522 Accepted Submission(s): 219
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
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.
1 2 3
2 3 4
2 4 5
2 5 6
#include <bits/stdc++.h> using namespace std; #define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-;
const double pi=acos(-1.0);
const int K=1e6+;
const int mod=1e9+; vector<PII>mp[K];
int n,k,sz[K];
LL ans;
void dfs(int u,int f)
{
sz[u]=;
for(auto v:mp[u])if(f!=v.first) dfs(v.first,u),sz[u]+=sz[v.first];
for(auto v:mp[u])if(f!=v.first) ans+=1LL*v.second*min(sz[v.first],k);
}
int main(void)
{ while(~scanf("%d%d",&n,&k))
{
ans=;
memset(mp,,sizeof mp);
for(int i=,x,y,z;i<n;i++)
scanf("%d%d%d",&x,&y,&z),mp[x].PB(MP(y,z)),mp[y].PB(MP(x,z));
dfs(,);
printf("%lld\n",ans);
}
return ;
}
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