动态规划训练之三

https://www.luogu.org/problem/P2515

读完题相信已经思路明确了

有依赖关系的连边，

但可能这整个图不连通，并且还有可能出现环

如果出现环的话，要选其中一个就必须吧整个环都选上（应该很好理解吧）

那么显然这要求我们进行tarjan缩点

缩点后是一个有向无环的森林，考虑建一个超级源点

对于每个入度为零的，与他相连

之后就是一个树上背包了（注意循环的顺倒序）

code：

#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]);
     //我想你会和我一样对这个转移式子很不解吧，为什么和平时写的不一样呢？
    //这个方程只有选当前这个点的情况,为什么?因为他们有依恋关系呀,只有选了当前的这个点才可以继续向下dp
   //再就是这个dp的是点权
    }
}
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
", (dp(num + 1), f[num + 1][m]));
    return 0;
}