poj 1947 Rebuilding Roads(樹形DP)

Rebuilding Roads Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 8227   Accepted: 3672 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. * 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.  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.  [A subtree with nodes (1, 2, 3, 6, 7, 8) will become isolated if roads 1-4 and 1-5 are destroyed.]  USACO 2002 February

#include <iostream>#include<stdio.h>#include<string.h>using namespace std;const int maxn=250;const int INF=0x3f3f3f3f;struct node{    int v;    node *next;} edge[maxn<<1],*head[maxn];int n,m,cnt,dp[maxn][maxn];void adde(int u,int v){    edge[cnt].v=v;    edge[cnt].next=head[u];    head[u]=&edge[cnt++];}void dfs(int fa,int u){    int i,j,v,mm=INF;    node *p;    dp[u][1]=0;    for(p=head[u];p!=NULL;p=p->next)    {        v=p->v;        if(v==fa)            continue;        dfs(u,v);        for(i=m;i>=1;i--)//01背包。從大到小裝        {            mm=dp[u][i]+1;//不要該子樹.            for(j=1;j<i;j++)//要該子樹                mm=min(mm,dp[u][j]+dp[v][i-j]);            dp[u][i]=mm;        }    }}int main(){    int i,u,v,ans;    while(~scanf("%d%d",&n,&m))    {        cnt=0;        memset(head,0,sizeof head);        memset(dp,0x3f,sizeof dp);        for(i=1;i<n;i++)        {            scanf("%d%d",&u,&v);            adde(u,v);            adde(v,u);        }        dfs(1,1);        ans=dp[1][m];        for(i=1;i<=n;i++)            ans=min(ans,dp[i][m]+1);//加1是因為根轉移了        printf("%d\n",ans);    }    return 0;}