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Cities

For each case, output one line, contain one integer, the minimum expect distance multiply K2.

1
2 2
1 2 1

2

易证树上所取子图连通

令dp[i,j]表示子树i取j个节点

则对u树的子树v  u-v 所取次数=2*子树v所取节点数*(K-子树v所取节点数)  K同题意

那么它满足 最优子结构

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<vector>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define F (100000007)
#define MAXN (2000+10)
#define MAXK (100+10)
#define MAXc (100000+10) 
#define INF (1000000000000LL)
typedef long long ll;

int n,k;
ll dp[MAXN][MAXK];
vector<int> edge[MAXN],dis[MAXN];
void dfs(int u,int fa)
{
	int sz=edge[u].size();
	Rep(i,sz)
	{
		int v=edge[u][i];
		ll d=dis[u][i];
		if (v!=fa)
		{
			dfs(v,u);
			ForD(j,k)
			{
				ForD(t,j-1) //至少取1个节点 
					dp[u][j]=min(dp[u][j],dp[u][t]+dp[v][j-t]+(j-t)*(k-(j-t))*d*2);
			}
			
			
		}
	}
}
int main()
{
//	freopen("1003.in","r",stdin);
//	freopen(".out","w",stdout);
	int T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&k);
		For(i,n) dis[i].resize(0);
		For(i,n) edge[i].resize(0);
		For(i,n-1)
		{
			int a,b,c;
			scanf("%d%d%d",&a,&b,&c);
			edge[a].push_back(b);
			edge[b].push_back(a);
			dis[a].push_back(c);
			dis[b].push_back(c);
		}
		
		MEM(dp)
		For(i,n)
			Fork(j,2,k)
			{
				dp[i][j]=(ll)INF;
			}
		
		dfs(1,-1);
		ll ans=INF;
		For(i,n) ans=min(ans,dp[i][k]);
		printf("%I64d\n",ans);	
	}
	
		
	return 0;
}

HDU 5148(Cities-树上背包)

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原文地址：http://blog.csdn.net/nike0good/article/details/45899561