标签:dp
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6000 Accepted Submission(s): 3548
3 2 0 1 0 2 0 3 7 4 2 2 0 1 0 4 2 1 7 1 7 6 2 2 0 0
5 13
/*
题意:中文题目
思路:树形dp+背包,把没有限制的点当成父亲节点,走完了这个点再走别的点
dp[v][i] 走到v节点走了i个点的价值
转移方程
dp[v][i]=max(dp[v][i],dp[v][i-j]+dp[to][j]);
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>
#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)
#define eps 1e-8
typedef __int64 ll;
using namespace std;
#define INF 0x3f3f3f3f
#define N 205
int va[N],n,p;
vector<int>g[N];
int dp[N][N];
int vis[N];
void dfs(int x)
{
int i,j;
vis[x]=1;
dp[x][1]=va[x];
for(i=0;i<g[x].size();i++)
{
int to=g[x][i];
if(!vis[to]) dfs(to);
for(int v=p;v>0;v--)
for(int j=1;j<v;j++)
dp[x][v]=max(dp[x][v],dp[x][v-j]+dp[to][j]);
}
}
int main()
{
int i,j;
while(scanf("%d%d",&n,&p),n+p)
{
p++;
for(i=0;i<=n;i++)
g[i].clear();
int x;
for(i=1;i<=n;i++)
{
scanf("%d%d",&x,&va[i]);
g[x].push_back(i);
}
memset(vis,0,sizeof(vis));
memset(dp,0,sizeof(dp));
dfs(0);
printf("%d\n",dp[0][p]);
}
return 0;
}
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HDU 1561 The more, The Better(树形dp+背包)
标签:dp
原文地址:http://blog.csdn.net/u014737310/article/details/46798775