bzoj3784 树上的路径

传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=3784

【题解】

和超级钢琴很像啊。

一看题目,无脑点分。

那么我们发现点分的路径形式和“超级钢琴”的哪个很像啊。

点分是子树合并对吧,当前子树内的每个点都能选择前面子树的区间,那么跟“超级钢琴”不同的就是这个是加的。

那么点分治维护即可。距离似乎(?)不会太多,那么我们套用那个堆的做法就行。

upd:距离最多nlogn个。

# include <queue> 
# include <stdio.h>
# include <string.h>
# include <algorithm>
// # include <bits/stdc++.h>
 
using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e5 + 10, N = 1e6 + 10;
const int mod = 1e9+7;

# define RG register

int n, m;
int head[M], nxt[M], to[M], tot=0, w[M];
inline void add(int u, int v, int _w) {
    ++tot; nxt[tot] = head[u]; head[u] = tot; to[tot] = v; w[tot] = _w; 
}
inline void adde(int u, int v, int _w) {
    add(u, v, _w), add(v, u, _w);
}

int val[N], L[N], R[N], tn = 0; 

namespace DFZ {
    bool vis[M]; 
    int sz[M], mx[M];
    inline void getsize(int x, int fa=0) {
        sz[x] = 1, mx[x] = 0;
        for (int i=head[x]; i; i=nxt[i]) {
            if(to[i] == fa || vis[to[i]]) continue;
            getsize(to[i], x);
            sz[x] += sz[to[i]];
            if(sz[to[i]] > mx[x]) mx[x] = sz[to[i]]; 
        }
    }
    int mi, centre;
    inline void getcentre(int x, int tp, int fa=0) {
        if(sz[tp] - sz[x] > mx[x]) mx[x] = sz[tp] - sz[x];
        if(mx[x] < mi) mi = mx[x], centre = x;
        for (int i=head[x]; i; i=nxt[i]) {
            if(to[i] == fa || vis[to[i]]) continue;
            getcentre(to[i], tp, x);
        }
    }
    int curl, curr;
    
    inline void calc(int x, int fa, int dis) {
        val[++tn] = dis, L[tn] = curl, R[tn] = curr; 
        for (int i=head[x]; i; i=nxt[i]) 
            if(to[i] != fa && !vis[to[i]]) calc(to[i], x, dis+w[i]);
    }
    
    inline void dfs(int x) {
        getsize(x); mi=n; getcentre(x, x);
        x = centre;
        
        ++tn;
        curl = curr = tn;
        
        for (int i=head[x]; i; i=nxt[i]) 
            if(!vis[to[i]]) calc(to[i], x, w[i]), curr = tn; 
        
        vis[x] = 1;
        for (int i=head[x]; i; i=nxt[i]) 
            if(!vis[to[i]]) dfs(to[i]); 
    }
}

namespace ST {
    struct rmq {
        int i, v;
        rmq() {}
        rmq(int i, int v) : i(i), v(v) {}
        friend bool operator <(rmq a, rmq b) {
            return a.v<b.v;
        }
        friend bool operator >(rmq a, rmq b) {
            return a.v>b.v;
        }
    };
    int n, Log2[N];
    rmq f[N][21];
    inline void init(int _n) {
        n = _n;
        Log2[1] = 0;
        for (int i=2; i<=n; ++i) Log2[i] = Log2[i>>1] + 1;
        for (int i=1; i<=n; ++i) f[i][0] = rmq(i, val[i]);
        for (int j=1; j<=20; ++j)
            for (int i=1; i+(1<<j)-1<=n; ++i) 
                 f[i][j] = max(f[i][j-1], f[i+(1<<(j-1))][j-1]);
    }
    inline bool getpos(int l, int r, int &i, int &v) {
        if(l > r) return 0;
        int len = Log2[r-l+1];
        rmq t = max(f[l][len], f[r-(1<<len)+1][len]);
        i = t.i, v = t.v;
        return 1;
    }
}


struct pa {
    int x, l, r, mx, v; 
    pa() {}
    pa(int x, int l, int r, int mx, int v) : x(x), l(l), r(r), mx(mx), v(v) {}
    friend bool operator < (pa a, pa b) {
        return a.v<b.v;
    }    
};

priority_queue<pa> q;

int main() {
    scanf("%d%d", &n, &m); 
    for (int i=1, u, v, _w; i<n; ++i) {
        scanf("%d%d%d", &u, &v, &_w);
        adde(u, v, _w);
    }
    DFZ::dfs(1); 
    ST::init(tn); 
    while(!q.empty()) q.pop();
    for (int i=1; i<=tn; ++i) {
        int va, po;
        ST::getpos(L[i], R[i], po, va);
        q.push(pa(i, L[i], R[i], po, va+val[i])); 
    }
    for (int i=1; i<=m; ++i) {
        pa t = q.top(); 
        printf("%d\n", t.v); 
        q.pop(); 
        int va, po;
        if(ST::getpos(t.l, t.mx-1, po, va)) 
            q.push(pa(t.x, t.l, t.mx-1, po, val[t.x]+va));
        if(ST::getpos(t.mx+1, t.r, po, va))
            q.push(pa(t.x, t.mx+1, t.r, po, val[t.x]+va));
    }
    return 0;
}
View Code

 

posted @ 2017-05-01 20:09  Galaxies  阅读(115)  评论(0编辑  收藏