luoguP3714 [BJOI2017]树的难题 点分治

以后传数组绝对用指针...

 

考虑点分治

在点分的时候，把相同的颜色的在一起合并

之后，把不同颜色依次合并

我们可以用单调队列做到单次合并$O(n + m)$

如果我们按照深度大小来合并，那么由于每次都是把大的往小的去合并

因此，合并$n$的序列最多需要$2n$的势能

因此，最终我们就能达到$O(n \log n)$的统计复杂度

然而还有排序，所以实际复杂度$O(n \log^2 n)$，排序常数很小，自然能过

#include <set>
#include 
#include 
#include 
#include 
#include 
namespace remoon {
    #define re register
    #define de double
    #define le long double
    #define ri register int
    #define ll long long
    #define sh short
    #define pii pair
    #define mp make_pair
    #define pb push_back
    #define fi first
    #define se second
    #define tpr template 
    #define rep(iu, st, ed) for(ri iu = st; iu <= ed; iu ++)
    #define drep(iu, ed, st) for(ri iu = ed; iu >= st; iu --)    
    #define gc getchar
    inline int read() {
        int p = 0, w = 1; char c = gc();
        while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); }
        while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc();
        
        return p * w;
    }
    int wr[50], rw;
    #define pc(iw) putchar(iw)
    tpr inline void write(ra o, char c = '\n') {
        if(!o) pc('0');
        if(o < 0) o = -o, pc('-');
        while(o) wr[++ rw] = o % 10, o /= 10;
        while(rw) pc(wr[rw --] + '0');
        pc(c);
    }
    tpr inline void cmin(ra &a, ra b) { if(a > b) a = b; }
    tpr inline void cmax(ra &a, ra b) { if(a < b) a = b; } 
    tpr inline bool ckmin(ra &a, ra b) { return (a > b) ? a = b, 1 : 0; }
    tpr inline bool ckmax(ra &a, ra b) { return (a < b) ? a = b, 1 : 0; }
}
using namespace std;
using namespace remoon;

#define sid 400050
#define inf 1e9

int n, m, l, r, rt, asz, cnp, ans = -2e9;
int over[sid], son[sid], sz[sid], dep[sid], col[sid];
int cap[sid], nxt[sid], node[sid], cv[sid];

inline void addedge(int u, int v, int c) {
    nxt[++ cnp] = cap[u]; cap[u] = cnp;
    node[cnp] = v; col[cnp] = c;
}

struct Ans { 
    int f[sid], end;
} nowc, prec, now;

inline void init(Ans &a) { 
    rep(i, 0, a.end) a.f[i] = -inf; 
    a.end = 0;
}

inline void upd(Ans &a, Ans &b) {
    int len = max(a.end, b.end); a.end = len;
    rep(i, 0, len) {
        cmax(a.f[i], b.f[i]);
        if(l <= i && i <= r) cmax(ans, a.f[i]);
    }
}

int q[sid];
inline void qry(Ans &a, Ans &b, int opt = 0) {
    if(l == 1 && r == n - 1) {
        int mx1 = -inf, mx2 = -inf;
        rep(i, 0, a.end) cmax(mx1, a.f[i]);
        rep(j, 0, b.end) cmax(mx2, b.f[j]);
        cmax(ans, mx1 + mx2 - opt);
    }
    else {
        int fr = 1, to = 0;
        drep(i, min(r, b.end), l) {
            while(fr <= to && b.f[i] >= b.f[q[to]]) to --;
            q[++ to] = i;
        }
        rep(i, 0, a.end) {
            if(l - i >= 0) {
                while(fr <= to && b.f[l - i] >= b.f[q[to]]) to --;
                q[++ to] = l - i;
            }
            while(fr <= to && q[fr] > r - i) fr ++;
            if(fr <= to) cmax(ans, a.f[i] + b.f[q[fr]] - opt);
        }
    }
}

int vis[sid], tim;
vector  all, c[sid];

#define cur node[i]
inline void grt(int o, int fa) {
    sz[o] = 1; son[o] = 0; 
    for(int i = cap[o]; i; i = nxt[i]) 
    if(!over[cur] && cur != fa){
        grt(cur, o); sz[o] += sz[cur];
        cmax(son[o], sz[cur]);
    }
    cmax(son[o], asz - sz[o]);
    if(son[o] < son[rt]) rt = o;
}

inline int gh(int o, int fa) {
    int tmp = dep[o];
    for(ri i = cap[o]; i; i = nxt[i])
    if(cur != fa && !over[cur]) {
        dep[cur] = dep[o] + 1;
        cmax(tmp, gh(cur, o));
    }
    return tmp;
}

inline void gt(int o, int fa, int val, int lst) {
    sz[o] = 1;
    for(int i = cap[o]; i; i = nxt[i])
    if(!over[cur] && cur != fa) {
        int C = col[i];
        int nxt = (C == lst) ? 0 : cv[C];
        gt(cur, o, val + nxt, C);
        sz[o] += sz[cur];
    }
    cmax(now.end, dep[o]);
    cmax(now.f[dep[o]], val);
}

inline void solve(int o) {    

    over[o] = 1; ++ tim;
    for(int i = cap[o]; i; i = nxt[i]) 
    if(!over[cur]) {
        int C = col[i];
        if(vis[C] != tim) c[C].clear();
        vis[C] = tim; dep[cur] = 1;
        c[C].pb(mp(gh(cur, o), cur));
    }
    ++ tim;
    all.clear();
    for(int i = cap[o]; i; i = nxt[i])
    if(!over[cur]) {
        int C = col[i];
        if(vis[C] != tim) {
            vis[C] = tim; 
            sort(c[C].begin(), c[C].end());
            int len = c[C][c[C].size() - 1].fi;
            all.pb(mp(len, C));
        }
    }
    sort(all.begin(), all.end());
    
    init(prec);
    for(auto x : all) {
        int C = x.se; init(nowc);
        for(auto y : c[C]) {
            init(now); gt(y.se, o, cv[C], C); 
            qry(now, nowc, cv[C]); upd(nowc, now);
        }
        qry(prec, nowc); upd(prec, nowc);
    }
    
    for(int i = cap[o]; i; i = nxt[i])
    if(!over[cur]) {
        asz = sz[cur]; rt = 0;
        grt(cur, o); solve(rt);
    }
}

int main() {
    n = read(); m = read(); 
    l = read(); r = read();
    rep(i, 1, m) cv[i] = read();
    rep(i, 2, n) {
        int u = read(), v = read(), w = read();
        addedge(u, v, w); addedge(v, u, w);
    }
    rep(i, 0, n) now.f[i] = -inf;
    rep(i, 0, n) prec.f[i] = -inf;
    rep(i, 0, n) nowc.f[i] = -inf;
    asz = n; son[0] = n;
    grt(1, 0); solve(rt);
    write(ans);
    return 0;
}

 

转载于:https://www.cnblogs.com/reverymoon/p/9799047.html