Rebuilding Roads （树形dp）

最新推荐文章于 2020-09-12 21:51:58 发布

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#动态规划 #树形dp #算法

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在上一次地震后，奶牛们重建了农夫约翰的农场，其中包括150座谷仓（1 <= N <= 150），每个谷仓都有唯一的路径到达其他谷仓，形成了一个树形结构。约翰想了解如果发生另一场地震，破坏最少的哪些道路能导致P（1 <= P <= N）座谷仓与其余谷仓隔离。

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

Sample Output

Hint

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

题目大概：

由n个节点组成的树，问最少需要删除几条边，才能形成一个p个节点组成的树。

思路：

dp[i][j]是第i个结点包含j个节点组成一棵树需要删除的边数。

开始要初始化 dp[i][1]=f[i].第i个结点的孩子数+1。这个1是和父节点的联系。

然后遍历树，并利用初始化进行dp

状态转移方程

dp[x][j]=min(dp[x][j],dp[x][j-k]+dp[y][k]-2); 这里减去2是因为父亲和孩子之间有两个联系。因为之前加了一，

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;

vector<int>f[160];
int dp[160][160];
int n,p;
int son[160];

void dfs(int x)
{
    int l=f[x].size();
    for(int i=0;i<l;i++)
    {
        int y=f[x][i];
        dfs(y);
        for(int j=p;j>1;j--)
        {
            for(int k=1;k<j;k++)
            {
                dp[x][j]=min(dp[x][j],dp[x][j-k]+dp[y][k]-2);
            }
        }
    }
}
int main()
{
    scanf("%d%d",&n,&p);
    memset(son,0,sizeof(son));
    for(int i=1;i<=n;i++)f[i].clear();
    for(int i=1;i<n;i++)
    {
        int q,w;
        scanf("%d%d",&q,&w);
        f[q].push_back(w);
        son[w]=1;
    }
    int root=1;
    while(son[root])
    {
        root++;
    }
    for(int i=1;i<=n;i++)
    {
        dp[i][1]=f[i].size()+1;
        for(int j=2;j<=p;j++)
        {
            dp[i][j]=10000;
        }
    }
    dfs(root);
    dp[root][p]--;
    int ans=10000;
    for(int i=1;i<=n;i++)
    {
        ans=min(ans,dp[i][p]);
    }
    printf("%d\n",ans);


    return 0;
}