GYM 101350 F. Monkeying Around

F. Monkeying Around

time limit per test

2.0 s

memory limit per test

256 MB

input

standard input

output

standard output

When the monkey professor leaves his class for a short time, all the monkeys go bananas. N monkeys are lined up sitting side by side on their chairs. They each have the same joke book. Before the professor returns, M jokes were heard.

Each of the M jokes are said in the order given and have the following properties:

x_i - position of the monkey who said it.

l_i – index of the joke in the book.

k_i – volume the monkey says that joke.

When the monkey at position x_i says the joke l_i, all monkeys at a distance less than or equal to k_i from that monkey (including the monkey who said the joke) will fall off their chairs in laughter if they have never heard the joke l_i before.

If the joke l_i has been heard anytime during the past before, and the monkey hears it again, then he will sit back up in his chair.

A monkey can fall off his chair more than once (every time he hears a new joke), and if he is already on the ground and hears a new joke, he will stay on the ground.

Can you figure out how many monkeys will be in their seats by the time the professor comes back?

Input

The first line of input is T – the number of test cases.

The first line of each test case is N, M (1 ≤ N ≤ 10⁵) (1 ≤ M ≤ 10⁵) – the number of monkeys in the class, and the number of jokes said before the professor returns.

The next M lines contain the description of each joke: x_i, l_i, k_i (1 ≤ x_i ≤ N) (1 ≤ l_i ≤ 10⁵) (0 ≤ k_i ≤ N).

Output

For each test case, output on a line a single integer - the number of monkeys in their seats after all jokes have been said.

Example

Input

1
10 7
3 11 0
3 11 2
5 12 1
8 13 2
7 11 2
10 12 1
9 12 0

Output

3

【题解】

考虑   每一个人  站着还是坐下  取决于  最后一个他听到的笑话
这个可以线段树nlogn求出
然后
我们依次考虑每一个笑话
看他覆盖的区间
此笑话决定的人  那个人被覆盖0或>1次则坐下  否则站起来
也可以线段树
综上 复杂度nlogn

（这是我跟某dalao的聊天记录直接放这里了）

网上有用扫描线做的，不是很明白。。。

待我理解理解

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))

inline void swap(long long &x, long long &y)
{
    long long tmp = x;x = y;y = tmp;
}

inline void read(long long &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9')c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
    if(c == '-')x = -x;
}

const long long INF = 0x3f3f3f3f;
const long long MAXN = 1000000 + 10;

struct Node
{
    long long x, l, k, t;
}node[MAXN];

long long n, m, l[MAXN], r[MAXN], tot, cnt[MAXN];


//第一颗线段树，用来维护最晚时间颜色coloe 
long long color[MAXN], num[MAXN];

void pushup1(long long o, long long l, long long r)
{
    long long mid = (l + r) >> 1;
    if(!color[o << 1]) color[o << 1] = color[o];
    if(!color[o << 1 | 1]) color[o << 1 | 1] = color[o]; 
}

void modify1(long long ll, long long rr,long long col, long long o = 1, long long l = 1, long long r = n)
{
    pushup1(o, l, r);
    if(color[o]) return;
    if(ll <= l && rr >= r)
    {
        if(!color[o])color[o] = col;
        pushup1(o, l, r);
        return;
    }
    long long mid = (l + r) >> 1;
    if(mid >= ll)modify1(ll, rr, col, o << 1, l, mid);
    if(mid < rr) modify1(ll, rr, col, o << 1 | 1, mid + 1, r);
} 

void build1(long long o = 1, long long l = 1, long long r = n)
{
    if(l == r)
    {
        num[l] = color[o];
        return;
    }
    pushup1(o, l, r);
    long long mid = (l + r) >> 1;
    build1(o << 1, l, mid);
    build1(o << 1 | 1, mid + 1, r);
}

//第二颗线段树，维护特定颜色作用于某个点多少次 
long long data[MAXN], lazy[MAXN];

void pushup2(long long o, long long l, long long r)
{
    if(lazy[o])
    {
        long long mid = (l + r) >> 1;
        lazy[o << 1] += lazy[o];
        lazy[o << 1 | 1] += lazy[o];
        data[o << 1] += lazy[o] * (mid - l + 1);
        data[o << 1 | 1] += lazy[o] * (r - mid);
        lazy[o] = 0;
    }
}

void modify2(long long ll, long long rr, long long x, long long o = 1, long long l = 1, long long r = n)
{
    pushup2(o, l, r);
    if(ll <= l && rr >= r)
    {
        lazy[o] += x;
        data[o] += (r - l + 1) * x;
        return;
    }
    long long mid = (l + r) >> 1;
    if(mid >= ll)modify2(ll, rr, x, o << 1, l, mid);
    if(mid < rr) modify2(ll, rr, x, o << 1 | 1, mid + 1, r);
    data[o] = data[o << 1] + data[o << 1 | 1];
}

long long ask2(long long p, long long o = 1, long long l = 1, long long r = n)
{
    if(l == r && l == p)
        return data[o];
    pushup2(o, l, r);
    long long mid = (l + r) >> 1;
    if(p <= mid)return ask2(p, o << 1, l, mid);
    else return ask2(p, o << 1 | 1, mid + 1, r);
}

long long cmppp(Node a, Node b)
{
    return a.l < b.l; 
}

long long cmp(Node a, Node b)
{
    return a.t > b.t; 
}

long long cmpp(long long a, long long b)
{
    return num[a] < num[b];
}

long long ans, t;

int main()
{
//    freopen("data.txt", "r", stdin);
/*long long n, m,cnt[MAXN];
*/
    read(t);
    for(;t;-- t)
    {
        ans = 0;tot = 0;
        memset(color, 0, sizeof(color));
        memset(num, 0, sizeof(num));
        memset(data, 0, sizeof(data));
        memset(lazy, 0, sizeof(lazy));
        memset(l, 0, sizeof(l));
        memset(r, 0, sizeof(r));
        memset(node, 0, sizeof(node));
        memset(cnt, 0, sizeof(cnt));
        read(n), read(m);
        for(register long long i = 1;i <= m;++ i)
        read(node[i].x), read(node[i].l), read(node[i].k), node[i].t = i;
        for(register long long i = 1;i <= n;++ i) cnt[i] = i;
        for(register long long i = m;i >= 1;-- i)
            modify1(node[i].x - node[i].k, min(n, node[i].x + node[i].k), node[i].l);
        build1();
        std::sort(node + 1, node + 1 + m, cmppp);
        std::sort(cnt + 1, cnt + 1 + n, cmpp);
        long long now = node[1].l;tot = 1;l[1] = 1;
        for(register long long i = 2;i <= m;++ i)
            if(now != node[i].l)r[tot] = i - 1, ++ tot, now = node[i].l, l[tot] = i;
        r[tot] = m;
        for(register long long i = 1, p = 1;i <= tot;++ i)
        {
            for(register long long j = l[i];j <= r[i];++ j)
                modify2(node[j].x - node[j].k, min(n, node[j].x + node[j].k), 1);
            long long tmp = 0, flag = p;
            while(num[cnt[p]] == node[l[i]].l)
            {
                long long tmp = ask2(cnt[p]);
                if(tmp == 0 || tmp > 1) ++ ans;
                ++ p;
            }
            for(register long long j = l[i];j <= r[i];++ j)
                modify2(node[j].x - node[j].k, min(n, node[j].x + node[j].k), -1);
        }
        printf("%I64d
", ans);
    }
    return 0;
}