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树形DP [POJ 1947] Rebuilding Roads

2024-08-01 23:31:36   221人阅读

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 61 21 31 41 52 62 72 84 94 104 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
 
树形背包、- -
#include <iostream>#include <cstring>#include <cstdio>using namespace std;#define INF 0x7ffffff#define N 155struct 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++)              //除了根节点，其他节点要想成为独立的根，必先与父节点断绝关系，所以要先加1        {            ans=min(ans,dp[i][m]+1);        }        printf("%d\n",ans);    }        return 0;}

 

树形DP [POJ 1947] Rebuilding Roads


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