2016北京集训测试赛(十一)Problem C: 树链问题

Description

Solution

智障暴力题, 每个点维护一下子树信息, 树剖就好了. 我居然还傻了写了一发毛毛虫...

#include <cstdio>
#include <cctype>
#include <vector>
#include <algorithm>
#include <cstring>
#define vector std::vector
#define max std::max
#define min std::min
#define sort std::sort
#define swap std::swap

namespace Zeonfai
{
    inline int getInt()
    {
        int a = 0, sgn = 1; char c;
        while(! isdigit(c = getchar())) if(c == '-') sgn *= -1;
        while(isdigit(c)) a = a * 10 + c - '0', c = getchar();
        return a * sgn;
    }
}
const int N = (int)1e5;
int n, m;
int a[N];
struct query
{
    int u, v, val;
    inline query(int _u, int _v, int _val) {u = _u; v = _v; val = _val;}
};
struct binaryIndexedTree
{
    int a[N + 1];
    inline void clear() {memset(a, 0, sizeof(a));}
    inline void modify(int pos, int x)
    {
        for(int i = pos; i <= n; i += i & - i) a[i] += x;
    }
    inline int query(int pos)
    {
        int res = 0;
        for(int i = pos; i; i -= i & - i) res += a[i];
        return res;
    }
    inline int query(int L, int R)
    {
        if(L > R) return 0; else return query(R) - query(L - 1);
    }
}BIT[2];
struct tree
{
    struct node
    {
        vector<int> edg;
        int hvy, tp, pre;
        int sz, dep;
        int id[2], L, R;
        vector<query> qry;
        int ans;
        inline void clear() {edg.clear(); qry.clear();}
    }nd[N + 1];
    inline void clear() {for(int i = 1; i <= n; ++ i) nd[i].clear();}
    inline void addEdge(int u, int v) {nd[u].edg.push_back(v); nd[v].edg.push_back(u);}
    void getSize(int u, int pre)
    {
        nd[u].sz = 1; nd[u].dep = ~ pre ? nd[pre].dep + 1 : 0; nd[u].hvy = -1; nd[u].pre = pre;
        for(auto v : nd[u].edg) if(v != pre)
        {
            getSize(v, u); nd[u].sz += nd[v].sz;
            if(nd[u].hvy == -1 || nd[v].sz > nd[nd[u].hvy].sz) nd[u].hvy = v;
        }
    }
    int clk;
    void decomposition(int u, int tp)
    {
        nd[u].tp = tp; nd[u].id[0] = ++ clk;
        if(~ nd[u].hvy) decomposition(nd[u].hvy, tp);
        for(auto v : nd[u].edg) if(v != nd[u].pre && v != nd[u].hvy) decomposition(v, v);
    }
    inline void decomposition() {getSize(1, -1); clk = 0; decomposition(1, 1);}
    void getSection(int u)
    {
        if(nd[u].hvy == -1) return;
        nd[u].L = (int)1e9; nd[u].R = -1;
        for(auto v : nd[u].edg) if(v != nd[u].pre) getSection(v), nd[u].L = min(nd[u].L, nd[v].id[1]), nd[u].R = max(nd[u].R, nd[v].id[1]);
    }
    inline void getSection() {getSection(1);}
    inline int getLCA(int u, int v)
    {
        while(nd[u].tp != nd[v].tp)
        {
            if(nd[nd[u].tp].dep > nd[nd[v].tp].dep) u = nd[nd[u].tp].pre;
            else if(nd[nd[u].tp].dep < nd[nd[v].tp].dep) v = nd[nd[v].tp].pre;
            else u = nd[nd[u].tp].pre, v = nd[nd[v].tp].pre;
        }
        if(nd[u].dep > nd[v].dep) swap(u, v);
        return u;
    }
    inline int getPath(int u, int v, int opt)
    {
        int res = 0;
        if(opt)
        {
            if(nd[u].hvy == -1) u = nd[u].pre; if(nd[v].hvy == -1) v = nd[v].pre;
            while(nd[u].tp != nd[v].tp)
            {
                if(nd[nd[u].tp].dep > nd[nd[v].tp].dep) res += BIT[1].query(nd[nd[u].tp].L, nd[u].R), u = nd[nd[u].tp].pre;
                else if(nd[nd[u].tp].dep < nd[nd[v].tp].dep) res += BIT[1].query(nd[nd[v].tp].L, nd[v].R), v = nd[nd[v].tp].pre;
                else res += BIT[1].query(nd[nd[u].tp].L, nd[u].R) + BIT[1].query(nd[nd[v].tp].L, nd[v].R), u = nd[nd[u].tp].pre, v = nd[nd[v].tp].pre;
            }
            if(nd[u].dep > nd[v].dep) swap(u, v);
            res += BIT[1].query(nd[u].L, nd[v].R);
        }
        else
        {
            while(nd[u].tp != nd[v].tp)
            {
                if(nd[nd[u].tp].dep > nd[nd[v].tp].dep) res += BIT[0].query(nd[nd[u].tp].id[0], nd[u].id[0]), u = nd[nd[u].tp].pre;
                else if(nd[nd[u].tp].dep < nd[nd[v].tp].dep) res += BIT[0].query(nd[nd[v].tp].id[0], nd[v].id[0]), v = nd[nd[v].tp].pre;
                else res += BIT[0].query(nd[nd[u].tp].id[0], nd[u].id[0]) + BIT[0].query(nd[nd[v].tp].id[0], nd[v].id[0]), u = nd[nd[u].tp].pre, v = nd[nd[v].tp].pre;
            }
            if(nd[u].dep > nd[v].dep) swap(u, v);
            res += BIT[0].query(nd[u].id[0], nd[v].id[0]);
        }
        return res;
    }
    int getAnswer(int u)
    {
        nd[u].ans = 0;
        for(auto v : nd[u].edg) if(v != nd[u].pre) nd[u].ans += getAnswer(v);
        for(auto qry : nd[u].qry)
        {
            int cur = qry.val;
            cur += getPath(qry.u, qry.v, 1) - getPath(qry.u, qry.v, 0);
            nd[u].ans = max(nd[u].ans, cur);
        }
        for(int i = 0; i < 2; ++ i) BIT[i].modify(nd[u].id[i], nd[u].ans);
        return nd[u].ans;
    }
    inline int getAnswer() {return getAnswer(1);}
}T;
inline int cmp(int a, int b)
{
    int preA = T.nd[a].pre, preB = T.nd[b].pre;
    if(preA == -1) return 1; else if(preB == -1) return 0; else return T.nd[preA].id[0] < T.nd[preB].id[0];
}
int main()
{

    #ifndef ONLINE_JUDGE

    freopen("tchain.in", "r", stdin);
    freopen("tchain.out", "w", stdout);

    #endif

    using namespace Zeonfai;
    for(int cs = getInt(); cs --;)
    {
        n = getInt(), m = getInt();
        T.clear(); for(int i = 0; i < 2; ++ i) BIT[i].clear();
        for(int i = 1; i < n; ++ i)
        {
            int u = getInt(), v = getInt();
            T.addEdge(u, v);
        }
        T.decomposition();
        for(int i = 1; i <= n; ++ i) a[i] = i; sort(a + 1, a + n + 1, cmp); for(int i = 1; i <= n; ++ i) T.nd[a[i]].id[1] = i;
        T.getSection();
        for(int i = 0; i < m; ++ i)
        {
            int u = getInt(), v = getInt(), val = getInt();
            T.nd[T.getLCA(u, v)].qry.push_back(query(u, v, val));
        }
        printf("%d\n", T.getAnswer());
    }
}

posted @ 2017-08-21 16:40  Zeonfai  阅读(207)  评论(0编辑  收藏  举报