ZOJ 3626 Treasure Hunt I

刷水题，贴代码。比较裸的树形DP。。dp(i,j)表示以i节点为根节点用j天时间能够获得的最大价值。由于要往返，m除以2就可以了。

#include <iostream>
#include <cstdio>
#include <cstring>
#define maxn 101
using namespace std;
int n,m,start,e;
int dp[maxn][maxn],next[maxn << 1],v[maxn << 1],cost[maxn << 1],first[maxn],visit[maxn],val[maxn];

inline int max(int a,int b)
{
    return a > b ? a : b;
}
void add_edge(int a,int b,int c)
{
    cost[e] = c;
    v[e] = b;
    next[e] = first[a];
    first[a] = e;
    e++;
}

void read_graph()
{
    e = 0;
    memset(first,-1,sizeof(first));
    memset(visit,0,sizeof(visit));
    int i,j,a,b,c;
    for(i = 1;i <= n;i++)
        scanf("%d",&val[i]);

    for(i = 1;i < n;i++)
    {
        scanf("%d%d%d",&a,&b,&c);
        add_edge(a,b,c);
        add_edge(b,a,c);
    }
    scanf("%d%d",&start,&m);
    m /= 2;
    for(i = 1;i <= n;i++)
        for(j = 0;j <= m;j++)
            dp[i][j] = val[i];
}

void dfs(int u)
{
    visit[u] = 1;
    int i,j,k,x;
    for(i = first[u];i != -1;i = next[i])
    {
        x = v[i];
        if(visit[x])    continue;
        dfs(x);
        for(j = m;j >= cost[i];j--)
            for(k = 0;k <= j - cost[i];k++)
                dp[u][j] = max(dp[u][j],dp[u][j - cost[i] - k] + dp[x][k]);
    }
}
int main()
{
    while(scanf("%d",&n) == 1)
    {
        read_graph();
        dfs(start);
        printf("%d\n",dp[start][m]);
    }
    return 0;
}