Luogu3177 [HAOI2015]树上染色

Posted 2021-02-09 xunzhen

题目蓝链

Description

给定一棵有(n)个节点的树，初始全为白色。你要在里面找到(k)个点，并把它们染成黑色。要使得染完色后，黑点两两之间的距离加上白点两两之间的距离的和最大

Solution

我们可以设(dp[i][j])表示以(i)为根的子树中选择(j)个节点对全局答案的最大贡献

然后对于每一个非叶子节点，枚举儿子子树进行转移
[ dp[u][j + k] leftarrow dp[u][j] + dp[v][k] + val ]
其中(val)为以(v)为根的子树内的点通过对全局答案的贡献

至于时间复杂度，我们可以发现任意一对点只会在它们LCA处被计算一次，所以复杂度为(mathcal{O}(n^2))

Code

#include <bits/stdc++.h>

using namespace std;

#define fst first
#define snd second
#define mp make_pair
#define squ(x) ((LL)(x) * (x))
#define debug(...) fprintf(stderr, __VA_ARGS__)

typedef long long LL;
typedef pair<int, int> pii;

template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }

inline int read() {
    int sum = 0, fg = 1; char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == ‘-‘) fg = -1;
    for (; isdigit(c); c = getchar()) sum = (sum << 3) + (sum << 1) + (c ^ 0x30);
    return fg * sum;
}

const int maxn = 2e3 + 10;

int Begin[maxn], Next[maxn << 1], To[maxn << 1], w[maxn << 1], e;
inline void link(int x, int y, int z) { To[++e] = y, Next[e] = Begin[x], Begin[x] = e, w[e] = z; }

int n, m, sz[maxn];
LL dp[maxn][maxn];

inline void dfs(int now, int f) {
    int lim = 1;
    for (int i = Begin[now]; i; i = Next[i]) {
        int son = To[i];
        if (son == f) continue;
        dfs(son, now);
        for (int j = min(lim, m); ~j; j--)
            for (int k = min(sz[son], m); ~k; k--)
                if (j + k <= m)
                    chkmax(dp[now][j + k], dp[now][j] + dp[son][k] + (LL) w[i] * (k * (m - k) + (sz[son] - k) * (n - m - sz[son] + k)));
        lim += sz[son];
    }
    sz[now] = lim;
}

int main() {
#ifdef xunzhen
    freopen("tree.in", "r", stdin);
    freopen("tree.out", "w", stdout);
#endif

    n = read(), m = read();
    chkmin(m, n - m);
    for (int i = 1; i < n; i++) {
        int x = read(), y = read(), z = read();
        link(x, y, z), link(y, x, z);
    }

    dfs(1, 0);

    printf("%lld
", dp[1][m]);

    return 0;
}