[luoguP2948] [USACO09OPEN]滑雪课Ski Lessons（DP）

传送门

f[i][j]表示i时刻能力值为j的最大滑雪数

显然f[0][1]=0,开始搜索

三种转移：

①美美的喝上一杯**：f[i+1][j]=max(f[i+1][j],f[i][j])

②滑雪，f[i+当前能力值所能滑雪最短时间][j]=max(f[i+当前能力值所能滑雪最短时间][j],f[i][j])

③上课，对于所有i时刻开始的课，f[i+该课所需时间][该课达到能力值]=max(f[i+该课所需时间][该课达到能力值],f[i][j])

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 100001
#define max(x, y) ((x) > (y) ? (x) : (y))
#define min(x, y) ((x) < (y) ? (x) : (y))

int t, s, n, ans, mx = 1;
int val[N], f[N][101];
//val[i]表示滑雪能力为i时一次滑雪的最低耗时
//f[i][j]表示时间为i，滑雪能力为j所能滑的最多次数 

struct ovo
{
	int m, l, a;
}q[N];

struct qwq
{
	int c, d;
}p[N];

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - ‘0‘;
	return x * f;
}

inline bool cmp(qwq x, qwq y)
{
	return x.c < y.c;
}

int main()
{
	int i, j;
	t = read();
	s = read();
	n = read();
	for(i = 1; i <= s; i++)
	{
		q[i].m = read();
		q[i].l = read();
		q[i].a = read();
	}
	for(i = 1; i <= n; i++)
	{
		p[i].c = read();
		p[i].d = read();
	}
	std::sort(p + 1, p + n + 1, cmp);
	j = 1;
	val[0] = 1e9;
	for(i = 1; i <= 100; i++)
	{
		val[i] = val[i - 1];
		for(; i == p[j].c && j <= n; j++) val[i] = min(val[i], p[j].d);
	}
	memset(f, -1, sizeof(f));
	f[0][1] = 0;
	for(i = 0; i <= t; i++)
	{
		mx = -1;
		for(j = 1; j <= 100; j++)
		{
			f[i + 1][j] = max(f[i + 1][j], f[i][j]);
			if(f[i][j] != -1 && val[j] != 1e9)
			{
				mx = max(mx, f[i][j]);
				f[i + val[j]][j] = max(f[i + val[j]][j], f[i][j] + 1);
			}
		}
		for(j = 1; j <= s; j++)
			if(mx != -1 && i >= q[j].m)
				f[i + q[j].l][q[j].a] = max(f[i + q[j].l][q[j].a], mx);
	}
	for(i = 1; i <= 100; i++) ans = max(ans, f[t][i]);
	printf("%d\n", ans);
	return 0;
}