Rebuilding Roads
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 7940 | Accepted: 3534 |
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.
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.]
题目的意思：
给你一棵树，求通过求通过删除最少的边，使得剩余子树有给定的节点数。
d[i][j]表示以i为根的子树删除j个结点所需删除的最小边数。
需要注意的是所要求得到的剩余子树没有要求一定包括整棵树的根结点。
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
#define Min_(a,b) (a<b?a:b)
int N,P,flag,Min;
int d[160][160],s[160]; //s[i]存下以i为根的子树所拥有的最多结点数
bool vis[160];
vector<int> son[160];
void DFS(int x)
{
int i,j,k,v;
if(x==flag)
s[x]=0;
else
s[x]=1;
d[x][0]=0;
for(i=0;i<son[x].size();i++)
{
v=son[x][i];
DFS(v);
s[x]+=s[v];
for(j=N;j>=0;j--)
for(k=0;k<=j;k++)
{
if(d[x][j-k]!=-1 && d[v][k]!=-1)
{
if(d[x][j]==-1)
d[x][j]=d[x][j-k]+d[v][k];
else
d[x][j]=Min_(d[x][j],d[x][j-k]+d[v][k]);
}
}
}
if(x!=flag)
{
d[x][s[x]]=1;
if(s[x]>P)
Min=d[x][s[x]-P]+1<Min?d[x][s[x]-P]+1:Min;
else if(s[x]==P)
Min=1;
}
}
int main()
{
int i,u,v;
while(scanf("%d%d",&N,&P)==2)
{
memset(vis,0,sizeof(vis));
for(i=1;i<=N;i++)
son[i].clear();
for(i=1;i<N;i++)
{
scanf("%d%d",&u,&v);
son[u].push_back(v);
vis[v]=1;
}
memset(d,-1,sizeof(d));
Min=1000;
for(i=1;i<=N;i++)
{
if(!vis[i])
{
flag=i; //flag标记根节点
DFS(i);
}
}
if(d[flag][N-P]<Min)
Min=d[flag][N-P];
printf("%d\n",Min);
}
return 0;
}