POJ 2486 Apple Tree

Description

Input

Output

Sample Input

2 1 
0 11
1 2
3 2
0 1 2
1 2
1 3

Sample Output

11
2

Source

尽管这个题目是看的解题报告。可是还是最终过了。这题真是难为了我very very very long long time time

dp[a][b][0]:节点a 走b 步回到节点a

dp[a][b][1]: 节点a 走 b 步不回到节点a

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <queue>
#include <set>
#define N 210
#define M 110
using namespace std;
struct num
{
    int x,y,next;
}a[2*N];
int b[N],Top,n,m;
int dp[N][N][2],temp[N][N][2];
int main()
{
    //freopen("data.txt","r",stdin);
    void addeage(int x,int y);
    void dfs(int u,int fa);
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            int x;
            scanf("%d",&x);
            dp[i][0][0] = x;
            dp[i][0][1] = x;
        }
        memset(b,-1,sizeof(b));
        Top = 0;
        for(int i=1;i<=n-1;i++)
        {
            int x,y;
            scanf("%d %d",&x,&y);
            addeage(x,y);
            addeage(y,x);
        }
        dfs(1,-1);
        int ans = 0;
        for(int i=0;i<=m;i++)
        {
            ans = max(ans,dp[1][i][0]);
        }
        printf("%d\n",ans);
    }
    return 0;
}
void addeage(int x,int y)
{
    a[Top].y = y;
    a[Top].next = b[x];
    b[x] = Top++;
}
void dfs(int u,int fa)
{
    for(int i=b[u];i!=-1;i=a[i].next)
    {
        int v = a[i].y;
        if(v==fa)
        {
            continue;
        }
        dfs(v,u);
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<=m;j++)
            {
                temp[i][j][0] = dp[i][j][0];
                temp[i][j][1] = dp[i][j][1];
            }
        }
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=i;j++)
            {
                dp[u][i][0] = max(dp[u][i][0],temp[u][i-j][1]+temp[v][j-1][0]);
                if(j>=2)
                {
                    dp[u][i][0] = max(dp[u][i][0],temp[v][j-2][1]+temp[u][i-j][0]);
                    dp[u][i][1] = max(dp[u][i][1],temp[u][i-j][1]+temp[v][j-2][1]);
                }
            }
        }
    }
}