Rebuilding Roads
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 8479 | Accepted: 3795 |
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.
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.
* 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.]
给定一棵树，求分离出一颗p个节点的子树所需要的最少的破坏边数，
乍一看，没思路，想了好久，总算是理解了，一个题目可能有多种思考的方式，总要找到符合自己的思考方式，
用dp[i][j]表示分离出以i为根，j为大小的子树，显然dp[i][1]为i节点度数，然后就是背包合并的过程，每两个连通块合并时，
需要减去2，就是多统计的2，树形dp轻松搞定。
代码：
/* ***********************************************
Author :xianxingwuguan
Created Time :2014-2-4 16:25:33
File Name :1.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
const int maxn=160;
int head[maxn],tol,deg[maxn],dp[maxn][maxn],m;
struct node{
int next,to;
}edge[2*maxn];
void add(int u,int v){
edge[tol].to=v;
edge[tol].next=head[u];
head[u]=tol++;
}
void dfs(int u,int fa){
dp[u][1]=deg[u];
for(int i=head[u];i!=-1;i=edge[i].next){
int v=edge[i].to;
if(v==fa)continue;
dfs(v,u);
for(int j=m;j>1;j--)
for(int k=1;k<j;k++)
dp[u][j]=min(dp[u][j],dp[u][j-k]+dp[v][k]-2);
}
}
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
int i,j,k,n;
while(~scanf("%d%d",&n,&m)){
memset(head,-1,sizeof(head));
tol=0;
memset(deg,0,sizeof(deg));
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
dp[i][j]=INF;
for(i=1;i<n;i++){
scanf("%d%d",&j,&k);
deg[j]++;
deg[k]++;
add(j,k);
add(k,j);
}
dfs(1,-1);
int ans=INF;
for(i=1;i<=n;i++)
ans=min(ans,dp[i][m]);
printf("%d\n",ans);
}
return 0;
}