luogu P1046 陶陶摘苹果

二次联通门 : luoguP1046

/*
    这个题好难.....

    由苹果树可知
    这应该是个树结构的题
    所以很自然的想到了用树链剖分来搞一下
    连边 最后查询以1为根节点的子树的权值和...

    
    从前闲的没事写着玩...
*/
#include <cstdio>
#define Max 3300

void read (int &now)
{
    now = 0;
    char word = getchar ();
    while (word > '9' || word < '0')
        word = getchar ();
    while (word >= '0' && word <= '9')
    {
        now = now * 10 + word - '0';
        word = getchar ();
    }
}

static int Hight;

struct Edge
{
    int to;
    int next;
};

int Edge_Count;
int edge_list[Max];
Edge edge[Max];

inline void AddEdge (int from, int to)
{
    Edge_Count++;
    edge[Edge_Count].to = to;
    edge[Edge_Count].next = edge_list[from];
    edge_list[from] = Edge_Count;
    Edge_Count++;
    edge[Edge_Count].to = from;
    edge[Edge_Count].next = edge_list[to];
    edge_list[to] = Edge_Count;
}

struct Point
{
    int size;
    int deep;
    int dis;
    int father;
    int tree_number;
    int End;
};

struct Tree
{
    int l;
    int r;
    int dis;
    int Mid;
};

Tree tree[Max];
Point point[Max];
int Count;

void Dfs_1 (int now, int father)
{
    int pos = Count++;
    point[now].father = father;
    point[now].deep = point[father].deep + 1;
    for (int i = edge_list[now]; i; i = edge[i].next)
    {
        if (edge[i].to == father)
            continue;
        Dfs_1 (edge[i].to, now);
    }
    point[now].size = Count - pos;
}

int tree_dis[Max];

void Dfs_2 (int now, int chain)
{
    int pos = 0;
    point[now].tree_number = ++Count;
    tree_dis[Count] = point[now].dis;
    for (int i = edge_list[now]; i; i = edge[i].next)
    {
        if (point[edge[i].to].tree_number)
            continue;
        if (point[edge[i].to].size > point[pos].size)
            pos = edge[i].to;
    }
    if (pos)
        Dfs_2 (pos, chain);
    for (int i = edge_list[now]; i; i = edge[i].next)
    {
        if (point[edge[i].to].tree_number)
            continue;
        Dfs_2 (edge[i].to, edge[i].to);
    }
    point[now].End = Count;
}

void Tree_Build (int l, int r, int now)
{
    tree[now].l = l;
    tree[now].r = r;
    if (l == r)
    {
        tree[now].dis = tree_dis[++Count] <= (30 + Hight) ? 1 : 0;
        return ;
    }
    tree[now].Mid = (l + r) >> 1;
    Tree_Build (l, tree[now].Mid, now << 1);
    Tree_Build (tree[now].Mid + 1, r, now << 1 | 1);
    tree[now].dis = tree[now << 1].dis + tree[now << 1 | 1].dis;
}

int Tree_Query (int l, int r, int now)
{
    if (tree[now].l == l && tree[now].r == r)
        return tree[now].dis;
    if (r <= tree[now].Mid)
        return Tree_Query (l, r, now << 1);
    else if (l > tree[now].Mid)
        return Tree_Query (l, r, now << 1 | 1);
    else
        return Tree_Query (l, tree[now].Mid, now << 1) + Tree_Query (tree[now].Mid + 1, r, now << 1 | 1);
}

int main (int argc, char *argv[])
{
    for (int i = 1; i <= 10; i++)
        read (point[i].dis);
    read (Hight);
    for (int i = 1; i < 10; i++)
        AddEdge (i, i + 1);
    Count = 0;
    Dfs_1 (1, 1);
    Count = 0;
    Dfs_2 (1, 0);
    Count = 0;
    Tree_Build (1, 10, 1);
    printf ("%d", Tree_Query (point[1].tree_number, point[1].End, 1));
    return 0;
}