[BZOJ 4890][TJOI2017]城市

传送门

$ \color{green} {solution : }$

我们可以暴力枚举断边，然后 $ O(n) $ 的跑一次换根 $ dp $，然后复杂度是 $ O(n * n) $ 的

#include <bits/stdc++.h>
using namespace std;

const int maxn = 100010;

template <typename T> inline void G(T &x) {
    x = 0; char o; bool f = false;
    for ( ; !isdigit(o = getchar()); ) {
        if( o == '-') {
            f = true;
        }
    }
    for ( ; isdigit(o); o = getchar()) {
        x = (x << 1) + (x << 3) + (o & 15);
    }
    if( f) {
        x = ~x + 1;
    }
}

template <typename T> inline bool check_Max(T &x, const T &y) {
    return x < y ? x = y, false : true;
}

template <typename T> inline bool check_Min(T &x, const T &y) {
    return x > y ? x = y, false : true;
}

struct Edge {
    int v, w; Edge *to;
    Edge ( int v, int w, Edge *to) : v(v), w(w), to(to) {}
    Edge () {}
}*head[maxn], pool[maxn<<1], *pis = pool;

int di[maxn], ndi[maxn];

inline void dfs1(int u, int pre, int &Mx) {
    di[u] = ndi[u] = 0;
    for ( Edge *now = head[u]; now; now = now->to) if( now->v ^ pre) {
        dfs1(now->v, u, Mx);
        check_Max(ndi[u], di[now->v] + now->w);
        if(ndi[u] > di[u]) {
            swap(ndi[u], di[u]);
        }
    }
    check_Max(Mx, ndi[u] + di[u]);
}

inline void dfs2(int u, int pre, int &Mx) {
    check_Min(Mx, di[u]);
    for ( Edge *now = head[u]; now; now = now->to) if( now->v ^ pre) {
        if( di[u] == di[now->v] + now->w) {
            check_Max(ndi[now->v], ndi[u] + now->w);
        }
        else {
            check_Max(ndi[now->v], di[u] + now->w);
        }
        if( ndi[now->v] > di[now->v]) {
            swap(di[now->v], ndi[now->v]);
        }
        dfs2(now->v, u, Mx);
    }
}

int n;

struct line {
    int l, r, w;
}data[maxn];

int main() {
    G(n);
    for ( register int i = 1; i < n; ++ i) {
        G(data[i].l); G(data[i].r); G(data[i].w);
        int u = data[i].l, v = data[i].r, w = data[i].w;
        head[u] = new (pis ++) Edge(v, w, head[u]); head[v] = new (pis ++) Edge(u, w, head[v]);
    }
    int ans = 0x7fffffff;
    for ( register int i = 1; i < n; ++ i) {
        int mx1 = 0, mx2 = 0, ret = 0;
        dfs1(data[i].l, data[i].r, mx1); dfs1(data[i].r, data[i].l, mx2);
        check_Max(ret, mx1); check_Max(ret, mx2);
        int mx3 = di[data[i].l], mx4 = di[data[i].r];
        dfs2(data[i].l, data[i].r, mx3); dfs2(data[i].r, data[i].l, mx4);
        check_Max(ret, mx3 + mx4 + data[i].w);
        check_Min(ans, ret);
    }
    printf("%d\n",ans);
    return 0;
}

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