[lnsyoj1521/luoguP2292/HNOI2004] 打鼹鼠

题意

给定 \(n\) 个点 \((x_i,y_i)\) 和对应时间 \(time_i\)，求从任意点开始，每单位时间静止或四向移动，在 \(time_i\) 时停留的点数的最大值，保证 \(time_i\) 顺序输入

sol

线性 dp
记 \(f_i\) 表示停留在第 \(i\) 个点时，点数的最大值，则转移方程为

\[f_i=\max_{j=1}^i f_j+1(dist_{i,j}\le time_j - time_i) \]

注意数组初始化全为 \(1\)

代码

#include <iostream>
#include <algorithm>
#include <cstring>
#define x first 
#define y second 

using namespace std;
typedef pair<int, int> PII;
typedef pair<int, PII> PIP;

const int N = 10005;

int n, m;
int f[N];
PIP g[N];

int dist(PII a, PII b){
    return abs(a.x - b.x) + abs(a.y - b.y);
}

int main(){
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; i ++ ) scanf("%d%d%d", &g[i].x, &g[i].y.x, &g[i].y.y);
    
    int ans = 1;
    for (int i = 1; i <= m; i ++ ) f[i] = 1;
    for (int i = 2; i <= m; i ++ )
        for (int j = 1; j < i; j ++ )
            if (dist(g[i].y, g[j].y) <= g[i].x - g[j].x) f[i] = max(f[i], f[j] + 1), ans = max(ans, f[i]);

    printf("%d\n", ans);

    return 0;
}