[NOIP2018 TG D1T3]赛道修建

题目大意：$NOIP2018\;TG\;D1T3$

题解：题目要求最短的赛道的长度最大，可以想达到二分答案，接着就是一个显然的树形$DP$。

发现对于一个点，它子树中若有两条链接起来比要求的答案大，一定接起来成为一条路径，因为接起来答案一定加一，而传递上去的话不一定。然后对于一条链，一定是找可行的最短的链与它相接，把尽可能长的链传递上去。找最小的可行的链我使用了双向链表（复杂度$O(n)$，右端点总共最多向左移动$n$次，每次最多向右移动$1$次）

卡点：考场上写结束后删除节点后转移到下一个节点时，没有考虑到移动到的节点也被删除的情况（考场上我是真的傻）

C++ Code：

#include <cstdio>
#include <algorithm>
#include <vector>
#include <cctype>

namespace R {
    int x, ch;
    inline int read() {
        ch = getchar();
        while (isspace(ch)) ch = getchar();
        for (x = ch & 15, ch = getchar(); isdigit(ch); ch = getchar()) x = x * 10 + (ch & 15);
        return x;
    }
}
using R::read;

#define maxn 50010
const int TANG_Yx = 20040826;
inline int max(int a, int b) {return a > b ? a : b;}
int head[maxn], cnt;
struct Edge {
    int to, nxt, w;
} e[maxn << 1];
inline void add(int a, int b, int c) {
    e[++cnt] = (Edge) {b, head[a], c}; head[a] = cnt;
}

int n, m, sum, ans;
int k, f[maxn];
inline bool debug(int k) {return true;}
int pre[maxn], nxt[maxn];

std::vector<int> V[maxn];
int dfn[maxn], rnk[maxn], idx, fa[maxn];
int up[maxn];

void dfs1(int u, int fa = 0) {
    ::fa[u] = fa; rnk[u] = u;
    dfn[u] = ++idx;
    for (int i = head[u]; i; i = e[i].nxt) {
        int v = e[i].to;
        if (v != fa) {
            up[v] = e[i].w;
            dfs1(v, u);
        }
    }
}

inline void work(int u, int fa) {
    std::vector<int> &V = ::V[u];
    std::sort(V.begin(), V.end());
    int sz = V.size();
    while (sz && V[sz - 1] >= k) f[u]++, sz--;
    int l = 0, r = 1, rem = 0;
    if (sz > 0) {
        #define End sz
        #define Begin (sz + 1)
        for (register int i = 0; i < sz; i++) {
            pre[i] = i - 1;
            nxt[i] = i + 1;
        }
        nxt[Begin] = 0;
        pre[0] = Begin;
        nxt[sz - 1] = End;
        pre[End] = sz - 1;
        while (r < End && l < r) {
            while (nxt[r] < End && V[l] + V[r] < k) r = nxt[r];
            while (pre[r] > l && pre[r] != Begin && V[l] + V[pre[r]] >= k) r = pre[r];
            if (V[l] + V[r] >= k) {
                f[u]++;
                nxt[pre[l]] = nxt[l];
                pre[nxt[l]] = pre[l];
                nxt[pre[r]] = nxt[r];
                pre[nxt[r]] = pre[r];
                if (nxt[pre[l]] != End && pre[nxt[r]] != Begin && pre[nxt[r]] > nxt[pre[l]]) r = pre[nxt[r]];
                else r = nxt[r];
                l = nxt[pre[l]];
            } else l = nxt[l];
            if (l == r) r = nxt[r];
        }
        if (0 <= pre[End] && pre[End] < sz) rem = V[pre[End]];
        else rem = 0;
        #undef End
        #undef Begin
    }
    if (u != 1) {
        ::V[fa].push_back(rem + up[u]);
        f[fa] += f[u];
    }
}
inline bool check(int mid) {
    k = mid;
    for (register int i = 1; i <= n; i++) V[i].clear(), f[i] = 0;
    for (register int I = 1, i = rnk[I]; I <= n; i = rnk[++I]) {
        work(i, fa[i]);
    }
    return f[1] >= m;
}

namespace Work1 {
    int MAX, ans;
    void dfs(int u, int fa = 0, int dep = 0) {
        if (dep > MAX) {
            MAX = dep;
            ans = u;
        }
        for (int i = head[u]; i; i = e[i].nxt) {
            int v = e[i].to;
            if (v != fa) {
                dfs(v, u, dep + e[i].w);
            }
        }
    }
    int main() {
        MAX = 0;
        dfs(1);
        int x = ans;
        MAX = 0;
        dfs(x);
        printf("%d\n", MAX);
        return 0;
    }
}

namespace Work2 {
    int pre[maxn];
    void dfs(int u, int fa = 0) {
        for (int i = head[u]; i; i = e[i].nxt) {
            int v = e[i].to;
            if (v != fa) {
                pre[v] = pre[u] + e[i].w;
                dfs(v, u);
            }
        }
    }
    bool check(int mid) {
        int last = 0, res = 0;
        for (int i = 1; i <= n; i++) {
            if (pre[i] - last >= mid) {
                last = pre[i];
                res++;
            }
        }
        return res >= m;
    }
    int main() {
        dfs(1);
        int l = 1, r = sum / m, ans = 0;
        while (l <= r) {
            int mid = l + r >> 1;
            if (check(mid)) {
                l = mid + 1;
                ans = mid;
            } else r = mid - 1;
        }
        printf("%d\n", ans);
        return 0;
    }
}

inline bool cmp(int a, int b) {return dfn[a] > dfn[b];}
bool flag = true;
int main() {
    n = read(), m = read();
    for (int i = 1, a, b, c; i < n; i++) {
        a = read(), b = read(), c = read();
        add(a, b, c);
        add(b, a, c);
        if (a - b != 1 && b - a != 1) flag = false;
        sum += c;
    }
    if (m == 1) {
        return Work1::main();
    }
    if (flag) {
        return Work2::main();
    }
    dfs1(1);
    std::sort(rnk + 1, rnk + n + 1, cmp);
    int l = 1, r = sum / m;
    while (l <= r) {
        int mid = l + r >> 1;
        if (check(mid)) {
            l = mid + 1;
            ans = mid;
        } else r = mid - 1;
    }
    printf("%d\n", ans);
    return 0;
}