ZOJ 3626 Treasure Hunt I

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>

///dp[k][j]表示以k的根的树用j天最多可以得到多少宝藏
using namespace std;
const int MAXN = 105;
int e;
int n, k, m;
int first[MAXN], next[MAXN << 1], v[MAXN << 1];
bool vis[MAXN];
int t[MAXN][MAXN], val[MAXN], dp[MAXN][MAXN];

void addedge(int x, int y)
{
v[e] = y, next[e] = first[x];
first[x] = e ++;
}

void ReadGraph()
{
for(int i = 1; i <= n; i ++)
scanf("%d", &val[i]);
e = 0;
memset(vis, false, sizeof vis);
memset(first, -1, sizeof first);
for(int j = 0; j < n - 1; j ++)
{
int ui, vi, di;
scanf("%d%d%d", &ui, &vi, &di);
t[ui][vi] = t[vi][ui] = di;
addedge(ui, vi);
addedge(vi, ui);
}
scanf("%d%d", &k, &m);
m /= 2;
}

void dfs(int k)
{
int i, j, x, s;
vis[k] = true;
for(s = first[k]; s != -1; s = next[s])
{
x = v[s];
if(!vis[x])
{
dfs(x);
for(j = m; j >= t[k][x]; j --) //从k有足够的时间去x点
for(i = 0; i <= j - t[k][x]; i ++)
{
dp[k][j] = max(dp[k][j], dp[k][j - t[k][x] - i] + dp[x][i]);
}
}
}
}

int main()
{
while(scanf("%d", &n) == 1)
{
ReadGraph();
for(int i = 1; i <= n; i ++)
for(int j = 0; j <= m; j ++)
dp[i][j] = val[i];
dfs(k);
printf("%d\n", dp[k][m]);
}
return 0;
}