[BZOJ2427][HAOI2010]软件安装（Tarjan+树形&背包DP）

可以发现，软件的依赖关系可以形成环，对于一个环内的软件，要么都安装，要么都不安装。所以这里先Tarjan强连通分量缩点，构成一个新图。新图显然是一个森林。以下，都把新图的节点个数定为n，新图中第i个点（强连通分量）的总占用空间定为Wi，总价值定为Vi。
建立虚拟节点n+1，定Wn+1=Vn+1=0，并把虚拟节点作为根，连接到新图里所有入度为0的点，形成一棵树。
然后在树上做背包DP。设f[u][i]为在点u的子树内，选出的占用空间不大于i的最大价值。
对于每个u，先初始化：
∀0≤i<Wu,f[u][i]=0
∀Wu≤i≤m,f[u][i]=Vi
然后逆序枚举i从m−Wu到0，顺序枚举j从0到i，则得出转移：
f[u][i+Wu]=max(f[u][i+Wu],f[u][i+Wu−j]+f[v][j])
最后结果就是f[n+1][m]。
代码：

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;inline int read() {    int res = 0; bool bo = 0; char c;    while (((c = getchar()) < '0' || c > '9') && c != '-');    if (c == '-') bo = 1; else res = c - 48;    while ((c = getchar()) >= '0' && c <= '9')        res = (res << 3) + (res << 1) + (c - 48);    return bo ? ~res + 1 : res;}const int N = 105, M = 505;int n, m, W[N], V[N], f[N][M], ecnt, nxt[M], adj[N], go[M], top,sta[N], dfn[N], low[N], times, num, bel[N], cost[N], val[N], ecnt2,nxt2[M], adj2[N], go2[M], d[N];bool ins[N], G[N][N];void add_edge(int u, int v) {    nxt[++ecnt] = adj[u]; adj[u] = ecnt; go[ecnt] = v;}void add_edge2(int u, int v) {    nxt2[++ecnt2] = adj2[u]; adj2[u] = ecnt2; go2[ecnt2] = v;}void Tarjan(int u) {    dfn[u] = low[u] = ++times;    sta[++top] = u; ins[u] = 1;    for (int e = adj[u], v; e; e = nxt[e])        if (!dfn[v = go[e]]) {            Tarjan(v);            low[u] = min(low[u], low[v]);        }        else if (ins[v]) low[u] = min(low[u], dfn[v]);    if (dfn[u] == low[u]) {        int v; bel[u] = ++num; ins[u] = 0;        while (v = sta[top--], v != u) bel[v] = num, ins[v] = 0;    }}void dp(int u) {    int i, j;    for (i = cost[u]; i <= m; i++) f[u][i] = val[u];    for (int e = adj2[u], v; e; e = nxt2[e]) {        dp(v = go2[e]);        for (i = m - cost[u]; i >= 0; i--) for (j = 0; j <= i; j++)            f[u][i + cost[u]] = max(f[u][i + cost[u]],                f[u][i + cost[u] - j] + f[v][j]);    }}int main() {    int i, j, x; n = read(); m = read();    for (i = 1; i <= n; i++) W[i] = read();    for (i = 1; i <= n; i++) V[i] = read();    for (i = 1; i <= n; i++) if (x = read()) add_edge(x, i);    for (i = 1; i <= n; i++) if (!dfn[i]) Tarjan(i);    for (i = 1; i <= n; i++) {        cost[bel[i]] += W[i]; val[bel[i]] += V[i];        for (int e = adj[i]; e; e = nxt[e])            if (bel[i] != bel[go[e]]) G[bel[i]][bel[go[e]]] = 1,                d[bel[go[e]]]++;    }    for (i = 1; i <= num; i++) for (j = 1; j <= num; j++)        if (G[i][j]) add_edge2(i, j);    for (i = 1; i <= num; i++) if (!d[i])        add_edge2(num + 1, i);    printf("%d\n", (dp(num + 1), f[num + 1][m]));    return 0;}