67 lines
1.8 KiB
C++
67 lines
1.8 KiB
C++
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#include <bits/stdc++.h>
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using namespace std;
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/**
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8. 泛化物品
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）题目：
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在背包容量为v的背包问题中，泛化物品是一个定义域为0..v中的整数的函数h，当分配给它的费用为v时，能得到的价值就是h(v)。
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为了应用泛化物品思想，这里假设题目为：存在依赖关系树（简单起见，只有一个根节点），即有依赖的物品也可以被依赖，可结合原文7.3节来理解此句。
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2）输入：
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测试用例数
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物品数 背包容量 根节点编号
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第i个物品的ci wi di(依赖物品的编号，-1为不依赖其他物品)
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* @return
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*/
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#define maxN 100
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#define maxV 1000
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int n, v;
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int cnt = 0;
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int head[maxN];
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int wi[maxN], ci[maxN];
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int f[maxN][maxV];
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struct Edge {
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int v, next;
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} e[maxN - 1];
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void addEdge(int u, int v) {
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e[cnt].v = v;
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e[cnt].next = head[u];
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head[u] = cnt++;
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}
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void treeDP(int u) {
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for (int i = ci[u]; i <= v; i++) {
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f[u][i] = wi[u];
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}
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for (int i = head[u]; i != -1; i = e[i].next) {
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int curr = e[i].v;
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treeDP(curr);
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for (int j = v; j >= 0; j--) {
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for (int k = j - ci[u]; k >= 0; k--) {
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f[u][j] = max(f[u][j], f[u][j - k] + f[curr][k]);
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}
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}
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}
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}
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int main(void) {
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int cases, root;
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freopen("../8.txt", "r", stdin);
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cin >> cases;
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while (cases--) {
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cnt = 0;
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memset(head, -1, sizeof(head));
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memset(f, 0, sizeof(f));
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cin >> n >> v >> root;
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for (int i = 0; i < n; i++) {
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int di;
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cin >> ci[i] >> wi[i] >> di;
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addEdge(di, i);
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}
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treeDP(root);
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cout << f[root][v] << endl;
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}
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}
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