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\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P1273 有线电视网

P1273 有线电视网

记 $f _ {i, j}$ 表示当前在第 $i$ 个点,满足 $j$ 个用户的最大利益,转移方程易得。但是复杂度为 $n ^ 3$ ,不能通过此题,一个优化是,只枚举当前子树中叶子节点个数的情况。

查看代码
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#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 3e3 + 10, M = 3e3 + 10, inf = 3e3;
int n, m, g[N], f[N][N];
int idx, hd[N], nxt[M], edg[M], wt[M];
int dfs(int x)
{
if (x > n - m)
return 1;
int s = 0;
for (int i = hd[x]; ~i; i = nxt[i])
{
int mx = dfs(edg[i]);
for (int j = 0; j <= s; j++)
g[j] = f[x][j];
for (int j = 0; j <= s + mx; j++)
for (int k = 0; k <= s; k++)
f[x][j] = max(f[x][j], g[k] + f[edg[i]][j - k] - wt[i]);
s += mx;
}
return s;
}
void add(int a, int b, int c)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
wt[idx] = c;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
hd[i] = -1;
for (int i = 1, k; i <= n - m; i++)
{
scanf("%d", &k);
for (int a, c; k; k--)
{
scanf("%d%d", &a, &c);
add(i, a, c);
}
}
for (int i = 1; i <= n; i++)
{
f[i][0] = 0;
for (int j = 1; j <= m; j++)
f[i][j] = -inf;
}
for (int i = n - m + 1, s; i <= n; i++)
{
scanf("%d", &s);
f[i][1] = s;
}
dfs(1);
for (int i = m; i >= 0; i--)
if (f[1][i] >= 0)
{
printf("%d", i);
break;
}
return 0;
}