Solution

枚举每一个鼹鼠出现的位置，并且从前往后转移。看看你要敲打这个鼹鼠的前提下，还可以继续往前敲打几个鼹鼠。这就是一题简单动态规划的题目了。

我们使用代表一定敲击这个鼹鼠最多的得分，那么枚举每一个前面的就可以找到答案，取即可。

#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
#define rep(i, sta, en) for(int i=sta; i<=en; ++i)
#define repp(i, sta, en) for(int i=sta; i>=en; --i)
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x, int op = 10) { if (!x) { putchar('0'); if (op)    putchar(op); return; }    char F[40]; ll tmp = x > 0 ? x : -x;    if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]);    if (op)    putchar(op); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
struct Node {
    int t, x, y;
    bool operator < (const Node& A) {
        return t < A.t;
    }
}p[10007];

const int N = 1e4 + 7;
ll n, m;
int f[N];

int dis(Node A, Node B) {
    return abs(A.x - B.x) + abs(A.y - B.y);
}

void solve() {
    n = read(), m = read();
    rep(i, 1, m)    p[i].t = read(), p[i].x = read(), p[i].y = read();
    int ans = 1;
    rep(i, 1, m) {
        f[i] = 1;
        rep(j, 1, i - 1) {
            if (p[i].t - p[j].t >= dis(p[i], p[j]))
                f[i] = max(f[i], f[j] + 1);
        }
        ans = max(ans, f[i]);
    }
    print(ans);
}

int main() {
    //int T = read();    while (T--)
    solve();
    return 0;
}