GYM 101350 F. Monkeying Around(线段树 or 思维）

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

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我开始傻×的把操作存了起来，然后按笑话的种类排序，可是后来发现不能这么做啊，室友先后顺序的！！
因为每一只猴子的最后的状态要看他最后一次听到的笑话是什么，然后再看这个笑话听到了几次若果是==1，就从椅子下来。

第一步求每一只猴子最后听到那种笑话：首先可以线段树，然后也可以set求
第二部 求每一只猴子最后一次听到的笑话的种类的数量
有两种方法！
1： for每一只猴子，然后在猴子的位置上对每种笑话进行查分，num【i】 保存每种颜色的数量
2：for每一个笑话的种类，在for在这个种类上的猴子并查询

第一种码：

#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <string>  
#include <algorithm>  
#include <cmath>  
#include <map>  
#include <set>  
#include <stack>  
#include <queue>  
#include <vector>  
#include <bitset>  
#include <functional>  
 
using namespace std;
 
#define LL long long  
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 10;
 
int n, m, x, y, z;
int a[maxn << 2], lazy[maxn << 2];
int sum[maxn];
vector<pair<int, int> > g1[maxn];
vector<int>g2[maxn];
 
void build(int k, int l, int r)
{
    a[k] = lazy[k] = 0;
    if (l == r) return;
    int mid = (l + r) / 2;
    build(k << 1, l, mid);
    build(k << 1 | 1, mid + 1, r);
}
 
void pushDown(int k)
{
    a[k << 1] = a[k << 1 | 1] = lazy[k];
    lazy[k << 1] = lazy[k << 1 | 1] = lazy[k];
    lazy[k] = 0;
}
 
void update(int k, int l, int r, int ll, int rr, int val)
{
    if (ll <= l && r <= rr)
    {
        a[k] = lazy[k] = val;
        return;
    }
    int mid = (l + r) / 2;
    if (lazy[k]) pushDown(k);
    if (ll <= mid) update(k << 1, l, mid, ll, rr, val);
    if (rr > mid)  update(k << 1 | 1, mid + 1, r, ll, rr, val);
}
 
int get(int k, int l, int r, int p)
{
    if (l == r) return a[k];
    int mid = (l + r) / 2;
    if (lazy[k]) pushDown(k);
    if (p <= mid) return get(k << 1, l, mid, p);
    return get(k << 1 | 1, mid + 1, r, p);
}
 
int lowbit(int k)
{
    return k & -k;
}
 
void update(int k, int val)
{
    for (int i = k; i < maxn; i += lowbit(i))
        sum[i] += val;
}
 
int getsum(int k)
{
    int ans = 0;
    for (int i = k; i; i -= lowbit(i))
        ans = ans + sum[i];
    return ans;
}
 
int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d%d", &n, &m);
        for (int i = 0; i < maxn; i++)
        {
            g1[i].clear();
            g2[i].clear();
        }
        build(1, 1, n);
        for (int i = 1; i <= m; i++)
        {
            scanf("%d%d%d", &x, &y, &z);
            int L = max(1, x - z);
            int R = min(n, x + z);
            update(1, 1, n, L, R, y);
            g1[y].push_back(make_pair(L, R));
        }
        for (int i = 1; i <= n; i++)
        {
            int k = get(1, 1, n, i);
            g2[k].push_back(i);
        }
        int ans = g2[0].size();
        for (int i = 1; i < maxn; i++)
        {
            int Size1 = g1[i].size(), Size2 = g2[i].size();
            if (Size1 == 0 || Size2 == 0) continue;
            for (int j = 0; j < Size1; j++)
            {
                pair<int, int>p = g1[i][j];
                update(p.first, 1);
                update(p.second + 1, -1);
            }
            for (int j = 0; j < Size2; j++)
            {
                int p = g2[i][j];
                if (getsum(p) >= 2) ans++;
            }
            for (int j = 0; j < Size1; j++)
            {
                pair<int, int>p = g1[i][j];
                update(p.first, -1);
                update(p.second + 1, 1);
            }
        }
        printf("%d
", ans);
    }
    return 0;
}

第二种：

#include<cstdio>
#include<iostream>
#include<set>
#include<vector>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
const int maxn=100000+10;
set<int> s;
vector<int> G[maxn];
int n,m,num[maxn],type[maxn];
int main(void)
{
    int xx,ll,kk,T,ans;
    cin>>T;
    while(T--)
    {
        ans=0;
        for(int i=0;i<maxn;i++)
            G[i].clear();
        s.clear();
        memset(num,0,sizeof(num));
        scanf("%d%d",&n,&m);
        for(int ii=1;ii<=m;ii++)
        {
            scanf("%d%d%d",&xx,&ll,&kk);
            type[ii]=ll;
            int lend=max(1,xx-kk);
            int rend=min(n,xx+kk);
            G[lend].push_back(ii);
            G[rend+1].push_back(-ii);
        }
        for(int i=1;i<=n;i++)
        {
            for(auto it:G[i])
            {
                if(it>0)//This is the begin position for a joke
                {
                    s.insert(it);
                    num[type[it]]++;
                }
                else
                {
                    s.erase(-it);
                    num[type[-it]]--;
                }
            }
            if(s.empty())
                ans++;//important! No jokes at all
            else
            {
                auto it=s.end();
                it--;
                if(num[type[*it]]>1)
                    ans++;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}

#include<cstdio>
#include<iostream>
#include<set>
#include<vector>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
const int maxn=100000+10;
set<int> s;
vector<int> G[maxn];
int n,m,num[maxn],type[maxn];
int main(void)
{
    int xx,ll,kk,T,ans;
    cin>>T;
    while(T--)
    {
        ans=0;
        for(int i=0;i<maxn;i++)
            G[i].clear();
        s.clear();
        memset(num,0,sizeof(num));
        scanf("%d%d",&n,&m);
        for(int ii=1;ii<=m;ii++)
        {
            scanf("%d%d%d",&xx,&ll,&kk);
            type[ii]=ll;
            int lend=max(1,xx-kk);
            int rend=min(n,xx+kk);
            G[lend].push_back(ii);
            G[rend+1].push_back(-ii);
        }
        for(int i=1;i<=n;i++)
        {
            for(auto it:G[i])
            {
                if(it>0)//This is the begin position for a joke
                {
                    s.insert(it);
                    num[type[it]]++;
                }
                else
                {
                    s.erase(-it);
                    num[type[-it]]--;
                }
            }
            if(s.empty())
                ans++;//important! No jokes at all
            else
            {
                auto it=s.end();
                it--;
                if(num[type[*it]]>1)
                    ans++;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}