SPOJ 1825 Free tour II 树分治

题意：

给出一颗边带权的数，树上的点有黑色和白色。求一条长度最大且黑色节点不超过k个的最长路径，输出最长的长度。

分析：

说一下题目的坑点：

• 定义递归函数的前面要加inline，否则会RE。不知道这是什么鬼，=_=|。
• ans要初始化为0，而不是一个绝对值很大的负数，因为我们可以选择只有一个顶点的路径，这样权值就是0。

题解请戳这，还有这。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
#define MP make_pair
using namespace std;

typedef pair<int, int> PII;

const int maxn = 200000 + 10;
const int INF = 0x3f3f3f3f;

struct Edge
{
    int v, w, nxt;
    Edge() {}
    Edge(int v, int w, int nxt): v(v), w(w), nxt(nxt) {}
};

int n, k, m, ans;
int black[maxn];
vector<int> a;

int ecnt, head[maxn];
Edge edges[maxn * 2];

inline void AddEdge(int u, int v, int w) {
    edges[ecnt] = Edge(v, w, head[u]);
    head[u] = ecnt++;
}

int fa[maxn], sz[maxn];
int dep[maxn], f[maxn], g[maxn], num[maxn];
bool del[maxn];

inline void dfs(int u) {
    sz[u] = 1;
    for(int i = head[u]; ~i; i = edges[i].nxt) {
        int v = edges[i].v;
        if(del[v] || v == fa[u]) continue;
        fa[v] = u;
        dfs(v);
        sz[u] += sz[v];
    }
}

inline PII findCenter(int u, int t) {
    PII ans(INF, u);
    int m = 0;
    for(int i = head[u]; ~i; i = edges[i].nxt) {
        int v = edges[i].v;
        if(del[v] || v == fa[u]) continue;
        m = max(m, sz[v]);
        ans = min(ans, findCenter(v, t));
    }
    m = max(m, t - sz[u]);
    return min(ans, MP(m, u));
}

inline void getdep(int u, int p) {
    dep[u] = 0;
    for(int i = head[u]; ~i; i = edges[i].nxt) {
        int v = edges[i].v;
        if(v == p || del[v]) continue;
        getdep(v, u);
        dep[u] = max(dep[u], dep[v]);
    }
    dep[u] += black[u];
}

bool cmp(int a, int b) { return dep[edges[a].v] < dep[edges[b].v]; }

inline void getg(int u, int p, int d, int c) {
    g[c] = max(g[c], d);
    for(int i = head[u]; ~i; i = edges[i].nxt) {
        Edge& e = edges[i];
        if(e.v == p || del[e.v]) continue;
        getg(e.v, u, d + e.w, c + black[e.v]);
    }
}

inline void solve(int u) {
    fa[u] = 0;
    dfs(u);
    int s = findCenter(u, sz[u]).second;
    del[s] = true;
    
    for(int i = head[s]; ~i; i = edges[i].nxt) {
        int v = edges[i].v;
        if(del[v]) continue;
        solve(v);
    }

    int tot = 0;
    for(int i = head[s]; ~i; i = edges[i].nxt) {
        int v = edges[i].v;
        if(del[v]) continue;
        getdep(v, s);
        num[++tot] = i;
    }

    sort(num + 1, num + 1 + tot, cmp);

    int mxdep = dep[edges[num[tot]].v];
    for(int i = 0; i <= mxdep; i++) f[i] = -INF;
    for(int i = 1; i <= tot; i++) {
        int v = edges[num[i]].v, w = edges[num[i]].w;
        int d = dep[v];
        for(int j = 0; j <= d; j++) g[j] = -INF;
        getg(v, s, w, black[v]);

        if(i != 1) {
            for(int j = 0; j <= k - black[s] && j <= d; j++) {
                int j2 = min(dep[edges[num[i-1]].v], k - black[s] - j);
                if(f[j2] == -INF) break;
                if(g[j] != -INF) ans = max(ans, g[j] + f[j2]);
            }
        }

        for(int j = 0; j <= d; j++) {
            f[j] = max(f[j], g[j]);
            if(j) f[j] = max(f[j], f[j-1]);
            if(j + black[s] <= k) ans = max(ans, f[j]);
        }
    }

    del[s] = false;
}

int main()
{
    scanf("%d%d%d", &n, &k, &m);
    while(m--) {
        int x; scanf("%d", &x);
        black[x] = 1;
    }

    ecnt = 0;
    memset(head, -1, sizeof(head));
    for(int i = 1; i < n; i++) {
        int u, v, w; scanf("%d%d%d", &u, &v, &w);
        AddEdge(u, v, w);
        AddEdge(v, u, w);
    }

    ans = 0;
    solve(1);
    printf("%d\n", ans);

    return 0;
}