题目大意
树上依赖多重背包，允许从根出发的一条链中每个点上的一个物品免费，求最大价值。
题解
背包神仙题……
首先都能想到的就是枚举某个节点，这个节点到根上的所有节点使用一次免费的机会，那么整棵树就被分成了三部分：
1.这条链上所有点剩余物品数量为
；
2.这条链左边所有点不变。
3.这条链右边所有点不变。
于是这里可以按照dfs序来dp，
表示dp到了第i个节点，选了
个节点达到的最大价值。从子树更新上来的时候由于要强制必选一个点，因此每个点物品的上界应该是
而不是
，这样最后必选的那条链也刚好可以保证剩余物品数量为
。
接下来翻转所有节点挂的儿子顺序，再做一遍上面的dp，只是要在dp完所有儿子之后再计算自己的贡献。这个dp记为
，最后实际上就是求
。因此复杂度为
。
#include <bits/stdc++.h>
namespace IOStream {
const int MAXR = 10000000;
char _READ_[MAXR], _PRINT_[MAXR];
int _READ_POS_, _PRINT_POS_, _READ_LEN_;
inline char readc() {
#ifndef ONLINE_JUDGE
return getchar();
#endif
if (!_READ_POS_) _READ_LEN_ = fread(_READ_, 1, MAXR, stdin);
char c = _READ_[_READ_POS_++];
if (_READ_POS_ == MAXR) _READ_POS_ = 0;
if (_READ_POS_ > _READ_LEN_) return 0;
return c;
}
template<typename T> inline void read(T &x) {
x = 0; register int flag = 1, c;
while (((c = readc()) < '0' || c > '9') && c != '-');
if (c == '-') flag = -1; else x = c - '0';
while ((c = readc()) >= '0' && c <= '9') x = x * 10 + c - '0';
x *= flag;
}
template<typename T1, typename ...T2> inline void read(T1 &a, T2 &...x) {
read(a), read(x...);
}
inline int reads(char *s) {
register int len = 0, c;
while (isspace(c = readc()) || !c);
s[len++] = c;
while (!isspace(c = readc()) && c) s[len++] = c;
s[len] = 0;
return len;
}
inline void ioflush() {
fwrite(_PRINT_, 1, _PRINT_POS_, stdout), _PRINT_POS_ = 0;
fflush(stdout);
}
inline void printc(char c) {
_PRINT_[_PRINT_POS_++] = c;
if (_PRINT_POS_ == MAXR) ioflush();
}
inline void prints(char *s) {
for (int i = 0; s[i]; i++) printc(s[i]);
}
template<typename T> inline void print(T x, char c = '\n') {
if (x < 0) printc('-'), x = -x;
if (x) {
static char sta[20];
register int tp = 0;
for (; x; x /= 10) sta[tp++] = x % 10 + '0';
while (tp > 0) printc(sta[--tp]);
} else printc('0');
printc(c);
}
template<typename T1, typename ...T2> inline void print(T1 x, T2... y) {
print(x, ' '), print(y...);
}
}
using namespace IOStream;
using namespace std;
typedef long long ll;
const int MAXN = 20005, MAXK = 26000005, MAXM = 500005;
struct Edge { int to, next; } edge[MAXN];
int head[MAXN], lim[MAXN], val[MAXN], sum[MAXN], temp[MAXM], tot, n, m, T;
int poolf[MAXK], poolg[MAXK], *(f[MAXN]), *(g[MAXN]), par[MAXN];
void addedge(int u, int v) {
edge[++tot] = (Edge) { v, head[u] };
head[u] = tot;
}
int que[MAXM], ans;
void dfs1(int u, int fa) {
sum[u] = sum[fa] + val[u];
int *he = que + 1, *ta = que;
for (int i = 0; i <= m; i++) {
int now = f[fa][i] - i * val[u];
while (ta >= he && f[fa][*ta] - *ta * val[u] <= now) --ta;
*(++ta) = i;
if (*he + lim[u] == i) ++he;
f[u][i] = f[fa][*he] + (i - *he) * val[u];
}
for (int i = head[u]; i; i = edge[i].next) {
int v = edge[i].to;
dfs1(v, u);
for (int j = 1; j <= m; j++) f[u][j] = max(f[u][j], f[v][j - 1] + val[v]);
}
}
void dfs2(int u, int fa) {
for (int i = 0; i <= m; i++) g[u][i] = g[fa][i];
for (int i = head[u]; i; i = edge[i].next) {
int v = edge[i].to;
dfs2(v, u);
for (int j = 1; j <= m; j++) g[u][j] = max(g[u][j], g[v][j - 1] + val[v]);
}
if (!head[u]) for (int i = 0; i <= m; i++)
ans = max(ans, f[u][i] + sum[u] + g[u][m - i]);
for (int i = 0; i <= m; i++) temp[i] = g[u][i];
int *he = que + 1, *ta = que;
for (int i = 0; i <= m; i++) {
int now = temp[i] - i * val[u];
while (ta >= he && temp[*ta] - *ta * val[u] <= now) --ta;
*(++ta) = i;
if (*he + lim[u] == i) ++he;
g[u][i] = temp[*he] + (i - *he) * val[u];
}
}
int main() {
for (read(T); T--;) {
read(n, m);
memset(head, 0, sizeof(head)); tot = ans = 0;
for (int i = 1; i <= n; i++) {
read(par[i], lim[i], val[i]);
addedge(par[i], i);
}
memset(poolf, 0, sizeof(poolf));
memset(poolg, 0, sizeof(poolg));
f[0] = poolf, g[0] = poolg;
for (int i = 1; i <= n; i++) {
f[i] = f[i - 1] + m + 1;
g[i] = g[i - 1] + m + 1;
}
dfs1(1, 0);
memset(head, 0, sizeof(head)); tot = 0;
for (int i = n; i > 0; i--) addedge(par[i], i);
dfs2(1, 0);
print(ans);
}
ioflush();
return 0;
}