bzoj2427 软件安装

软件安装

题目背景：

bzoj2427

分析：树型DP

原图有环·····并且好像有和其他部分不连通的单个环······

这个题，显然应该是一些环和一些树放到一起，那么显然的对于一个环要么全部选择，要么一个都不选，那么我们先tarjan缩点一发每一个SCC的w为原来所有w的和，v为原来所有的v和，然后就成了一片森林，对于每棵树的根，我们向它连接一个虚拟根节点，然后直接树型DP即可，定义f[i][j]为，对于i和i的子树，话费j的最优价值。

注意：因为依存关系的存在，如果父节点不选择是不能选择子节点的，所以可以先将父节点的已有权值提出来，直接DP剩下的即可。

错误：知道吗，tarjan第二行的判断是，if (vis[v])不是vis[cur]······

Source：

/*created by scarlyw*/#include <cstdio>#include <string>#include <algorithm>#include <cstring>#include <iostream>#include <cmath>#include <cctype>#include <vector>#include <set>#include <queue>const int MAXN = 100 + 10;const int MAXM = 500 + 10;int n, m, o, ind, top;int w[MAXN], v[MAXN], d[MAXN], f[MAXN][MAXM], val[MAXN], sw[MAXN];int low[MAXN], num[MAXN], scc[MAXN], scnt[MAXN];std::vector<int> ori_edge[MAXN], edge[MAXN], s;bool vis[MAXN], map[MAXN][MAXN];inline void add_ori_edge(int x, int y) {ori_edge[x].push_back(y);}inline void read_in() {scanf("%d%d", &n, &m);for (int i = 1; i <= n; ++i) scanf("%d", &w[i]);for (int i = 1; i <= n; ++i) scanf("%d", &v[i]);for (int i = 1; i <= n; ++i) {scanf("%d", &d[i]);if (d[i]) add_ori_edge(d[i], i);}}inline void tarjan(int cur) {low[cur] = num[cur] = ++ind, vis[cur] = true, s.push_back(cur);for (int p = 0; p < ori_edge[cur].size(); ++p) {int v = ori_edge[cur][p];if (num[v] == 0) tarjan(v), low[cur] = std::min(low[cur], low[v]);else if (vis[v]) low[cur] = std::min(low[cur], num[v]);}if (low[cur] == num[cur]) {++top;do {o = s.back(), scc[o] = top, vis[o] = false;s.pop_back(), sw[top] += w[o], val[top] += v[o];} while (o != cur);}}inline void dfs(int cur) {for (int p = 0; p < edge[cur].size(); ++p) {int v = edge[cur][p];dfs(v);for (int k = m - sw[cur]; k >= 0; --k)for (int r = 0; r <= k; ++r)f[cur][k] = std::max(f[cur][k], f[cur][k - r] + f[v][r]);}for (int i = m; i >= sw[cur]; --i) f[cur][i] = f[cur][i - sw[cur]] + val[cur];for (int i = 0; i < sw[cur]; ++i) f[cur][i] = 0;}inline void solve() {for (int i = 1; i <= n; ++i)if (num[i] == 0) tarjan(i);for (int i = 0; i <= n; ++i) {for (int p = 0; p < ori_edge[i].size(); ++p) {int v = ori_edge[i][p];if (scc[i] != scc[v] && !map[scc[i]][scc[v]]) {edge[scc[i]].push_back(scc[v]), map[scc[i]][scc[v]] = true;vis[scc[v]] = true;}}}for (int i = 1; i <= top; ++i) if (!vis[i]) edge[0].push_back(i);dfs(0), std::cout << f[0][m];}int main() {read_in();solve();return 0;}