bzoj 1065(NOI 2008 奧運物流)

bzoj 1065(NOI 2008 奧運物流)

Description

自己看吧= =

Solution

這個題感覺很蛋疼啊= =因為它不僅僅是一棵樹，在1節點處還有一個環。我們考慮一個環上距離節點1距離為dep對答案所做的貢獻

假設環的長度為l,則貢獻為c1?kdep?(1+kl+k2l+....)，然後我們驚訝的發現這原來是個等比數列求和啊。。。

於是對答案的貢獻是c1?kdep1?kl,顯然，修改m次的話我們最好把每個節點都指向1，這樣dep=1，收益最大

之前一直沒想明白環怎麼辦，看別人題解發現直接一個等比數列求和就搞定了= =too young。。

然後用f[i][j][k]表示節點i為根的子樹，深度為j，修改k次獲得最大值，先枚舉1處的環長度再樹dp即可，詳情看代碼

Code

#include 
using namespace std;
typedef long long LL;
#define pb push_back
#define mp make_pair
#define F first
#define S second
inline void read(int &t) {
    int f = 1;char c;
    while (c = getchar(), c < '0' || c > '9') if (c == '-') f = -1;
    t = c - '0';
    while (c = getchar(), c >= '0' && c <= '9') t = t * 10 + c - '0';
    t *= f;
}
const int N = 65;
const double inf = 1e10;
vector g[N];
int tot, n, m, fa[N], cir[N], num[N];
double k, p[N], c[N], f[N][N][N];
void gao(int u, int dep, int len) {
    for (int i = 0, v; i < g[u].size(); ++i) {
        v = g[u][i];
        gao(v, dep + 1, len);
    }
    for (int i = 0; i <= dep; ++i) {
        if (num[u] == len && i != 1)    continue;
        if (u != 1 && num[u] && num[u] <= len  && i != (len - (tot - dep))) continue;
        f[u][i][0] = c[u] * p[i] / (1.0 - p[len]);
        for (int j = 0, v; j < g[u].size(); ++j) {
            v = g[u][j];
            for (int k = m; k >= 0; --k) {
                if (!k) f[u][i][k] += f[v][i + 1][k];
                else {
                    double t = -inf;
                    for (int l = 0; l <= k; ++l)
                        f[u][i][k] = max(f[u][i][k], f[u][i][k - l] + max(f[v][i + 1][l], l ? f[v][1][l - 1] : -inf));
                    f[u][i][k] = t;
                }
            }
        }
    }
}
int main() {
    read(n), read(m);
    scanf("%lf", &k);
    p[0] = 1.0;
    for (int i = 1; i <= n; ++i)    p[i] = p[i - 1] * k;
    for (int i = 1; i <= n; ++i) {
        read(fa[i]);
        if (i != 1) g[fa[i]].pb(i);
    }
    for (int i = 1; i <= n; ++i)    scanf("%lf", &c[i]);
    cir[++tot] = 1, num[1] = 1;
    int now = fa[1];
    while (now != 1) {
        cir[++tot] = now;
        num[now] = tot;
        now = fa[now];
    }
    double ans = 0.0;www.Bkjia.com
    for (int i = tot; i >= 2; --i) {//enum circle length
        for (int j = 1; j <= n; ++j)
            for (int k = 0; k <= n; ++k)
                for (int l = 0; l <= n; ++l)
                    f[j][k][l] = -inf;
        gao(1, 0, i);
        ans = max(ans, f[1][0][m]);
    }
    printf("%.2lf\n", ans);
    return 0;
}