[POJ 1947] Rebuilding Roads

Rebuilding Roads
Time Limit: 1000MS
 
Memory Limit: 30000K
Total Submissions: 9249
 
Accepted: 4198
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

木のリュック、--

#include <iostream>

#include <cstring>

#include <cstdio>

using namespace std;

#define INF 0x7ffffff

#define N 155



struct Edge

{

    int to,next;

}edge[N*N/2];

int head[N],tot;



int n,m;

int f[N];

int dp[N][N];



void add(int x,int y)

{

    edge[tot].to=y;

    edge[tot].next=head[x];

    head[x]=tot++;

}



void dfs(int u)

{

    int i,j,k;

    for(j=0;j<=m;j++)

    {

        dp[u][j]=INF;

    }

    dp[u][1]=0;

    for(i=head[u];i!=-1;i=edge[i].next)

    {

        int v=edge[i].to;

        dfs(v);

        for(j=m;j>=0;j--)

        {

            for(k=0;k<j;k++)

            {

                if(k)

                    dp[u][j] = min(dp[u][j],dp[u][j-k]+dp[v][k]);

                else

                    dp[u][j] = dp[u][j]+1;

            }

        }

    }

}



void init()

{

    tot=0;

    memset(f,-1,sizeof(f));

    memset(head,-1,sizeof(head));

}

int main()

{

    int i;

    while(scanf("%d%d",&n,&m), n||m)

    {

        init();

        for(i=1;i<n;i++)

        {

            int a,b;

            scanf("%d%d",&a,&b);

            add(a,b);

            f[b]=a;

        }

        int root=1;

        while(f[root]!=-1)

        {

            root=f[root];

        }

        dfs(root);

        int ans=dp[root][m];

        for(i=1;i<=n;i++)              // ， ， ， 1

        {

            ans=min(ans,dp[i][m]+1);

        }

        printf("%d
",ans);

    }    

    return 0;

}