LibreOJ #6192. 「美团 CodeM 复赛」城市网络

二次联通门 : LibreOJ #6192. 「美团 CodeM 复赛」城市网络

/*
    LibreOJ #6192. 「美团 CodeM 复赛」城市网络

    二分 + 单调栈 
    
    离线操作 
    
    维护一个严格单调的栈
    对于一个询问的答案
    就是以当前价值为限的单调栈的大小
    二分实现即可
    
    (我不会说是我不会倍增而硬是乱搞的) 
*/
#include <cstdio>
#include <iostream>
#include <vector>
#define INF 1e9

const int BUF = 12312312;
char Buf[BUF], *buf = Buf;
using std :: max;
inline void read (int &now)
{
    for (now = 0; !isdigit (*buf); ++ buf);
    for (; isdigit (*buf); now = now * 10 + *buf - '0', ++ buf);
}

#define Max 100010

struct E { E *n; int to; };
E *list[Max], poor[Max << 1], *Ta = poor;

int key[Max], deep[Max];
struct Data { int x, y; };

inline void In (int u, int v)
{
    ++ Ta, Ta->to = v, Ta->n = list[u], list[u] = Ta;
    ++ Ta, Ta->to = u, Ta->n = list[v], list[v] = Ta;
}
Data Stack[Max];
int N, top, Answer[Max];

struct Q 
{ 
    int to, key, Id; 
    Q () {}
    Q (int x, int y, int z) 
    {
        to = x, key = y, Id = z;
    } 
};
std :: vector <Q> data[Max];

void Dfs (int now, int Father)
{
    int l = 0, r = top + 1, Mid, pos;
    for (; l + 1 < r; )
    {
        Mid = l + r >> 1;
        if (Stack[Mid].x > key[now]) l = Mid;
        else r = Mid;
    }
    int t = top; top = r; Data res = Stack[r];
    Stack[top].x = key[now], Stack[top].y = now;
    if (now == 7)
    {
        for (int i = 1; i <= top; ++ i)
            printf ("%d %d
", Stack[i].x, Stack[i].y);
            puts ("");
        
    }
    for (std :: vector <Q> :: iterator i = data[now].begin (); i != data[now].end (); ++ i)
    {
        int _l = 0, _r = top + 1, _Mid;
        for (; _l + 1 < _r; )
        {
            _Mid = (_l + _r + 1) >> 1;
            if (Stack[_Mid].x > (*i).key) _l = _Mid;
            else _r = _Mid;
        }
        int __l = 0, __r = top + 1, __Mid;
        for (; __l + 1 < __r; )
        {
            __Mid = (__l + __r + 1) >> 1;
            if (deep[Stack[__Mid].y] < deep[(*i).to]) 
                __l = __Mid;
            else __r = __Mid;
        }
        if (now == 7) printf ("

%d %d

", _l, __l);
        Answer[(*i).Id] = max (0, _l - __l);
    } 
    for (E *e = list[now]; e; e = e->n)
        if (e->to != Father)
        {
            deep[e->to] = deep[now] + 1;
            Dfs (e->to, now);
        }
    Stack[top] = res, top = t;
}

int Main ()
{
    freopen ("1.in", "r", stdin);
    fread (buf, 1, BUF, stdin); int x, y, M;
    int z; read (N), read (M); register int i, j;
    for (i = 1; i <= N; ++ i) read (key[i]);
    for (i = 1; i < N; ++ i)
    {
        read (x), read (y);
        In (x, y);
    }
    for (i = 1; i <= M; ++ i)
    {
        read (x), read (y), read (z);
        data[x].push_back (Q (y, z, i));
    }
    deep[1] = 0, Stack[0].x = INF, Stack[0].y = 0;
    for (Dfs (1, 0), i = 1; i <= M; ++ i)
        printf ("%d
", Answer[i]);
    return 0;
}

int ZlycerQan = Main ();
int main (int argc, char *argv[]) {;}