poj2486 Apple Tree (tree backpack packets dp +)

Topic link: https: //vjudge.net/problem/POJ-2486

Meaning of the questions: a tree right point, the starting point 1, after seeking the right to up to a maximum point and m edges.

Ideas:

　　Tree dp classic title. With 3-dimensional state, dp [u] [j] [0/1] represents the maximum value j take steps in the sub-tree u (back u / do not return to u). Obviously dp [u] [j] [1]> = dp [u] [j] [0], so dp [1] [m] [1] is the final answer.

　　Suppose v is a child node of u, k and v is the number of steps in the v as used, the transfer equation in three steps, as follows:

　　　　dp[u][j][0]=max(dp[u][j][0] , dp[u][j-k][0]+dp[v][k-2][0]) (k>=2)

　　　　dp[u][j][1]=max(dp[u][j][1] , dp[u][j-k][0]+dp[v][k-1][1]) (k>=1)

　　　　dp [u] [j] [1] = max (dp [u] [j] [1], dp [u] [jk] [1] + dp [v] [k-2] [0]) (k > = 2) (step 3 easy to forget)

AC Code:

#include<cstdio>
#include<algorithm>
using namespace std;

const int maxn=105;
const int maxm=205;
int n,m,cnt,a[maxn],head[maxn],dp[maxn][maxm][2];

struct node{
    int v,nex;
}edge[maxn<<1];

void adde(int u,int v){
    edge[++cnt].v=v;
    edge[cnt].nex=head[u];
    head[u]=cnt;
}

void dfs(int u,int fa){
    for(int i=head[u];i;i=edge[i].nex){
        int v=edge[i].v;
        if(v==fa) continue;
        dfs(v,u);
        for(int j=m;j>=1;--j)
            for(int k=1;k<=j;++k){
                if(k>=2) dp[u][j][0]=max(dp[u][j][0],dp[u][j-k][0]+dp[v][k-2][0]);
                if(k>=1) dp[u][j][1]=max(dp[u][j][1],dp[u][j-k][0]+dp[v][k-1][1]);
                if(k>=2) dp[u][j][1]=max(dp[u][j][1],dp[u][j-k][1]+dp[v][k-2][0]);
            }
    }
}

int main(){
    while(~scanf("%d%d",&n,&m)){
        cnt=0;
        for(int i=1;i<=n;++i)
            scanf("%d",&a[i]);
        for(int i=1;i<=n;++i){
            head[i]=0;
            for(int j=0;j<=m;++j)
                dp[i][j][0]=dp[i][j][1]=a[i];
        }
        for(int i=1;i<n;++i){
            int u,v;
            scanf("%d%d",&u,&v);
            adde(u,v);
            adde(v,u);
        }
        dfs(1,0);
        printf("%d\n",dp[1][m][1]);
    }
    return 0;
}