P8816 [CSP-J 2022] 上升点列 题解

题目传送门：luogu P8816 [CSP-J 2022] 上升点列

这是一道简单的 dp 题。

定义到达指两点的曼哈顿距离是 \(1\)。

我们考虑设状态 \(f(i,j)\) 表示考虑结尾是第 \(i\) 个点，增加了 \(j\) 个点的情况。

\(f(i,j) = \max \{f(i,j), f(u,j) + 1\}\) 点 \(u\) 满足 \(u\) 可以被点 \(i\) 到达。

\(f(i,j) = \max \{f(i,j), f(u,j - dis) + dis + 1\}\) 点 \(u\) 满足可以被点 \(i\) 增加 \(dis\) 个点之后到达，且 \(dis < j\)。

答案就是 \(\max \{f(i,j) + k - j\}\)。

轻松ac(bushi

代码（良心马蜂，无压行

点击查看代码

#include <bits/stdc++.h>

namespace hello_world_djh {
    template <typename T>
    T read() {
        T x = 0, f = 1;
        char ch = getchar();
        for (; ch < '0' || ch > '9'; ch = getchar())
            if (ch == '-')
                f = ~f + 1;
        for (; ch >= '0' && ch <= '9'; ch = getchar())
            x = (x << 3) + (x << 1) + (ch ^ '0');
        return x * f;
    }
    
    unsigned popcount(unsigned x) {
		x = (x & 0x55555555) + ((x >> 1) & 0x55555555);
		x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
		x = (x & 0x0F0F0F0F) + ((x >> 4) & 0x0F0F0F0F);
		x = (x & 0x00FF00FF) + ((x >> 8) & 0x00FF00FF);
		x = (x & 0x0000FFFF) + ((x >> 16) & 0x0000FFFF);
		return x;
	}
	
	const int N = 510, K = 110;
	int f[N][K], n = read<int>(), k = read<int>();
	std::pair <int, int> point[N];
	
	#define x first
	#define y second

    int main() {
//    	freopen("out.txt", "w", stdout);
    	for (int i = 1; i <= n; i++) {
    		point[i].x = read<int>();
    		point[i].y = read<int>();
    		f[i][0] = 1;
		}
		std::sort(point + 1, point + n + 1);
		for (int i = 2; i <= n; i++)
			for (int j = 0; j <= k; j++) {
				if (j - 1 >= 0)
					f[i][j] = std::max(f[i][j], f[i][j - 1]);
				for (int u = 1; u < i; u++) {
					auto a = point[i];
					auto b = point[u];
					if (a.x < b.x || a.y < b.y)
						continue;
					if (abs(a.x - b.x) + abs(a.y - b.y) == 1)
						f[i][j] = std::max(f[i][j],
						f[u][j] + (abs(a.x - b.x) + abs(a.y - b.y) == 1));
				}
				for (int u = 1; u < i; u++) {
					auto a = point[i];
					auto b = point[u];
					if (a.x < b.x || a.y < b.y)
						continue;
					int dis = (a.x - b.x) + (a.y - b.y) - 1;
					if (j >= dis)
						f[i][j] = std::max(f[i][j], f[u][j - dis] + dis + 1);
				}
			}
		int ans = -1;
		for (int i = 1; i <= n; i++)
			for (int j = 0; j <= k; j++)
				ans = std::max(ans, f[i][j] + k - j);
		printf("%d\n", ans);
		return 0;
    }
};

int main() {
    hello_world_djh::main();
    return 0;
}