POJ题目1947 Rebuilding Roads（树形dp）

Rebuilding Roads

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 9957		Accepted: 4537

Description

The cows have reconstructed Farmer John's farm, with its N barns (1 <= N <= 150, number 1..N) after the terrible earthquake last May. The cows didn't have time to rebuild any extra roads, so now there is exactly one way to get from any given barn to any other
barn. Thus, the farm transportation system can be represented as a tree. 



Farmer John wants to know how much damage another earthquake could do. He wants to know the minimum number of roads whose destruction would isolate a subtree of exactly P (1 <= P <= N) barns from the rest of the barns.

Input

* Line 1: Two integers, N and P 



* Lines 2..N: N-1 lines, each with two integers I and J. Node I is node J's parent in the tree of roads. 

Output

A single line containing the integer that is the minimum number of roads that need to be destroyed for a subtree of P nodes to be isolated. 

Sample Input

11 6
1 2
1 3
1 4
1 5
2 6
2 7
2 8
4 9
4 10
4 11

Sample Output

2

Hint

[A subtree with nodes (1, 2, 3, 6, 7, 8) will become isolated if roads 1-4 and 1-5 are destroyed.] 

Source

USACO 2002 February

题目大意：问一个数删掉最少条边变成一个仅仅有n个结点的子树

ac代码

#include<stdio.h>
#include<string.h>
#define min(a,b) (a>b?

b:a)
#define INF 0xfffffff
int dp[220][220];
int pre[220],head[220],vis[220],dig[220];
int n,p,cnt;
struct s
{
	int u,v,w,next;
}edge[220*2];
void add(int u,int v)
{
	edge[cnt].u=u;
	edge[cnt].v=v;
	edge[cnt].next=head[u];
	head[u]=cnt++;
}
void tree_dp(int u)
{
	int i,j,k;
	for(i=0;i<=p;i++)
	{
		dp[u][i]=INF;
	}
	dp[u][1]=0;
	for(i=head[u];i!=-1;i=edge[i].next)
	{
		int v=edge[i].v;
		tree_dp(v);
		for(k=p;k>=1;k--)
		{
			dp[u][k]=dp[u][k]+1;
			for(j=1;j<k;j++)
			{

				dp[u][k]=min(dp[u][k],dp[u][j]+dp[v][k-j]);
			}
		}
	}
}
int DP(int u)
{
	tree_dp(u);
	int ans=dp[u][p];
	int i;
	for(i=1;i<=n;i++)
	{
		ans=min(ans,dp[i][p]+1);
	//	printf("%d\n",dp[i][1]);
	}
	return ans;
}
int main()
{
	//int n,p;
	while(scanf("%d%d",&n,&p)!=EOF)
	{
		int i;
		memset(dig,0,sizeof(dig));
		memset(head,-1,sizeof(head));
		cnt=0;
		for(i=0;i<n-1;i++)
		{
			int a,b;
			scanf("%d%d",&a,&b);
			add(a,b);
			dig[b]++;
		}
		int root;
		for(i=1;i<=n;i++)
		{
			if(dig[i]==0)
				root=i;
		}
		printf("%d\n",DP(root));
	}
}