SP1825 FTOUR2 - Free tour II 点分治+启发式合并+未调完

题意翻译

给定一棵n个点的树，树上有m个黑点，求出一条路径，使得这条路径经过的黑点数小于等于k，且路径长度最大

Code:

#include <bits/stdc++.h>
using namespace std;
#define pr pair<int, int>
#define mp make_pair
const int maxn = 2000003;
const int inf = 1000000000;
void setIO(string a)
{
    string in = a + ".in", out = a + ".out";
    freopen(in.c_str(), "r", stdin);
}
vector <pr> vrr;
int edges, n, m, K, ans, mdep, tl, root = 0, sn;
int hd[maxn], to[maxn << 1], nex[maxn << 1], val[maxn << 1], siz[maxn];
int mk[maxn],  vis[maxn], mine[maxn], dis[maxn], tmp1[maxn], tmp2[maxn], f[maxn];
void add(int u, int v, int c)
{
    nex[++edges] = hd[u], hd[u] = edges, to[edges] = v, val[edges] = c;
}
void Getroot(int u, int ff)
{
    siz[u] = 1, f[u] = 0;
    for(int i = hd[u]; i ; i = nex[i])
    {
        int v = to[i];
        if(v == ff || vis[v]) continue;
        Getroot(v, u);
        siz[u] += siz[v], f[u] = max(f[u], siz[v]);
    }
    f[u] = max(sn - siz[u], f[u]);
    if(f[u] < f[root]) root = u;
}
void getmax(int x, int ff, int d1)
{
    siz[x]=1;
    mdep = max(mdep, d1 + mk[x]);
    for(int i = hd[x]; i ; i = nex[i])
    {
        int v = to[i];
        if(vis[v] || v == ff) continue;
        dis[v] = dis[x] + val[i];
        getmax(v, x, d1 + mk[x]);
        siz[x]+=siz[v];
    }
}
void getdis(int x, int ff, int d2)
{
    if(d2 > K) return ;
    tmp1[++tl] = dis[x], tmp2[tl] = d2 + mk[x];
    for(int i = hd[x]; i ; i = nex[i])
    {
        int v = to[i];
        if(v == ff || vis[v]) continue;
        getdis(v, x, d2 + mk[x]);
    }
}
void calc(int x)
{
    if(mk[x]) --K;
    vrr.clear();
    siz[x]=1;
    for(int i = hd[x]; i ; i = nex[i])
    {
        int v = to[i];
        if(vis[v]) continue;
        mdep = 0, dis[v] = val[i];
        getmax(v, x, 0);
        siz[x]+=siz[v];
        vrr.push_back(mp(mdep, v));
    }
    sort(vrr.begin(), vrr.end());
    tl = mine[0] = 0;
   //  printf("%d ::: ",x);
    for(int sz = vrr.size(), i = 0; i < sz; ++i)
    {
        int cur = vrr[i].second;
        int pdl = tl;
        getdis(cur, x, 0);
        for(int j = pdl + 1; j <= tl; ++j)
            if(K >= tmp2[j])
                ans = max(ans, mine[K - tmp2[j]] + tmp1[j]);
        // printf("%d ",vrr[i].second);
        // tmp2[j] :: 有tmp2[j]个黑子时的最大值.
        for(int j = pdl + 1; j <= tl; ++j) mine[tmp2[j]] = max(mine[tmp2[j]], tmp1[j]);
        for(int j = 1; j <= vrr[i].first; ++j) mine[j] = max(mine[j-1],mine[j]);
    }
    // printf("\n");
    if(vrr.size())
        for(int i=1;i<=vrr[vrr.size()-1].first;++i) mine[i] = -inf;
    if(mk[x]) ++K;
}
void solve(int u)
{
    vis[u] = 1;
    calc(u);
    for(int i=hd[u];i;i=nex[i])
    {
        int v=to[i];
        if(vis[v]) continue;
        root=0,sn=siz[v];
        Getroot(v,u);
        solve(root);
    }
}
int main()
{
   //  setIO("input");
    scanf("%d%d%d",&n, &K, &m);
    for(int i = 1,o; i <= m ; ++i)
    {
        scanf("%d",&o);
        mk[o] = 1;
    }
    for(int i = 1; i < n ; ++i)
    {
        int u, v, c;
        scanf("%d%d%d",&u, &v, &c);
        add(u, v, c), add(v, u, c);
    }
    for(int i=1;i<maxn;++i) mine[i]=-inf;
    sn = n, f[0] = inf;
    Getroot(1, 0); solve(root);
    printf("%d\n",ans);
    return 0;
}