dp 百题计划（11 / 100）

P3205 [HNOI2010]合唱队

区间 dp，考虑 $dp[l][r][0 / 1]$ 表示让 $[l, r]$ 这一区间符合并且最后是从左 / 右进来的。

转移的话，考虑 $[l, r]$ 这段区间，可能是 $[l, r - 1]$ 和 $[l + 1, r]$ 这两段区间转移来的。

而这俩段区间都可能是从左边进来的或者从右边进来的，符合条件就转移即可。

 int n, a[N], dp[1020][1020][2];
signed main() 
{
	n = read();
	for(int i = 1; i <= n; i++) a[i] = read();
	for(int i = 1; i <= n; i++) dp[i][i][0] = 1;
	for(int len = 2; len <= n; len++) {
		for(int l = 1; l <= n - len + 1; l++) {
			int r = l + len - 1;
			if(a[l] < a[l + 1]) dp[l][r][0] = (dp[l][r][0] + dp[l + 1][r][0]) % mod;
			if(a[l] < a[r]) dp[l][r][0] = (dp[l][r][0] + dp[l + 1][r][1]) % mod;
			if(a[r] > a[r - 1]) dp[l][r][1] = (dp[l][r][1] + dp[l][r - 1][1]) % mod;
			if(a[r] > a[l]) dp[l][r][1] = (dp[l][r][1] + dp[l][r - 1][0]) % mod;
		}
	}
	printf("%d\n", (dp[1][n][1] + dp[1][n][0]) % mod);
	return 0;
}

P4170 [CQOI2007]涂色

区间dp, 考虑 $dp[l][r]$ 表示将 $[l, r]$ 涂完的最小次数。

如果说 $l = r$ ，答案明显是 $1$ 。

如果说 $s[l] == s[r](l\ne r)$ 可以少涂一次 $l$ 或者少涂一次 $r$ 。

所以这部分的转移方程是：

d p [l] [r] = min (d p [l] [r - 1], d p [l + 1] [r])

$dp[l][r] = \min(dp[l][r - 1], dp[l + 1][r])$

如果说 $s[l] \ne s[r]$ ，枚举端点。

d p [l, r] = min (d p [l] [r], d p [l] [k] + d p [k + 1] [r])

$dp[l, r] = \min(dp[l][r], dp[l][k] + dp[k +1][r])$

 char s[N];
int n, dp[100][100];
 
signed main() 
{
  scanf("%s", s + 1);
  n = strlen(s + 1);
  memset(dp, 0x3f, sizeof dp);
  for(int i = 1; i <= n; i++) dp[i][i] = 1;
  for(int len = 2; len <= n; len++) {
  	for(int l = 1, r = l + len - 1; l <= n - len + 1; l++, r++) {
  		if(s[l] == s[r]) dp[l][r] = min(dp[l][r - 1], dp[l + 1][r]);
  		else
  			for(int k = l; k < r; k++) dp[l][r] = min(dp[l][r], dp[l][k] + dp[k + 1][r]);
  	}
  }
  printf("%d\n", dp[1][n]);
  return 0;
}

P3842 [TJOI2007]线段

设 $dp_{i, 0 / 1 }$ 表示走到第 $i$ 条线段， $0$ 是在做左端点结束的， $1$ 是在右端点结束的。

分类讨论，从上一行转移，注意要取绝对值。

 #include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e4 + 10;
 
inline int read() {
	int s = 0, f = 0; char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) f |= (ch == '-');
	for(; isdigit(ch); ch = getchar()) s = (s << 1) + (s << 3) + (ch ^ '0');
	return f ? -s : s;
}
 
int n, dp[MAXN][2], l[MAXN], r[MAXN];
 
int main() {
	n = read();
	for(int i = 1; i <= n; i++) l[i] = read(), r[i] = read();
	dp[1][0] = r[1] + (r[1] - l[1] + 1), dp[1][1] = r[1] - 1;
	for(int i = 2; i <= n; i++) {
		dp[i][0] = min(dp[i - 1][0] + abs(r[i] - l[i - 1]) + abs(r[i] - l[i] + 1), dp[i - 1][1] + abs(r[i] - r[i - 1]) + abs(r[i] - l[i] + 1));
		dp[i][1] = min(dp[i - 1][0] + abs(l[i] - l[i - 1]) + abs(r[i] - l[i] + 1), dp[i - 1][1] + abs(r[i - 1] - l[i]) + abs(r[i] - l[i] + 1));
	}
	return printf("%d\n", min(dp[n][0] + abs(n - l[n]), dp[n][1] + abs(n - r[n]))), 0;
}

ABC 267 D

一眼 dp，设 $dp_{i, j}$ 为到第 $i$ 个位置，选了 $j$ 个。

所以 $dp_{i, j} = \max( dp_{i-1,j,}dp_{i-1, j-1} + a[i] \times j)$

十分显然，但是注意初始化。

 int n, a[N], m;
int dp[2020][2020];
signed main() {
 
	n = read(), m = read();
	for (int i = 0; i <= n; i++) for (int j = 1; j <= n; j++) dp[i][j] = -inf;
	for (int i = 1; i <= n; i++) a[i] = read();
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= min(i, m); j++) {
			dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - 1] + a[i] * j);
		}
	}
	int Max = -inf;
	for (int i = 1; i <= n; i++) Max = max(Max, dp[i][m]);
	cout << Max;
	return (0 - 0);
}

P1107 [BJWC2008]雷涛的小猫

比较一眼的 dp 吧。

设 $dp_{i, j}$ 表示到了第 i 层高度，第 j 颗树的最大柿子数量。

d p_{i, j} = max {d p_{i - 1, j}, {max}_{i = 1}^{n} d p_{i - d e l, j}} + a_{i, j}

$dp_{i,j} = \max\{dp_{i - 1, j},\max_{i=1}^ndp_{i-del,j}\} + a_{i,j}$

然后可以用一个数组来维护最大值，就做到了 $\mathcal{O}(n^2)$ 的复杂度。

 /**
 *	author: TLE_Automation
 *	creater: 2022.10.14
 **/
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<bitset>
#include<cstring>
#include<iostream>
#include<algorithm>
#define gc getchar
using namespace std;
typedef long long ll;
const int N = 1e6 + 10;
const int Maxn = 2e3 + 10;
const int mod = 998244353;
const ll inf = 0x3f3f3f3f3f3f3f3f;
#define debug cout << "i ak ioi" << "\n"
inline void print(int x) {if (x < 0) putchar('-'), x = -x; if(x > 9) print(x / 10); putchar(x % 10 + '0');}
inline char readchar() {static char buf[100000], *p1 = buf, *p2 = buf; return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;}
inline int read() { int res = 0, f = 0; char ch = gc();for (; !isdigit(ch); ch = gc()) f |= (ch == '-'); for (; isdigit(ch); ch = gc()) res = (res << 1) + (res << 3) + (ch ^ '0'); return f ? -res : res;}
 
int n, h, del, a[Maxn][Maxn], dp[Maxn][Maxn], Max[Maxn];
 
/*
	dp[i][j] 表示到了第 i 层高度，第 j 颗树的最大柿子数量
*/
 
signed main()
{
	n = read(), h = read(), del = read();
	for(int i = 1; i <= n; i++) {
		int x = read();
		for(int j = 1; j <= x; j++) a[read()][i]++;
	}
	for(int i = 1; i <= h; i++) {
		for(int j = 1; j <= n; j++) {
			if(i > del) dp[i][j] = std::max(dp[i - 1][j], Max[i - del]) + a[i][j];
			else dp[i][j] = dp[i - 1][j] + a[i][j];
			Max[i] = std::max(Max[i], dp[i][j]);
		}
	}
	int ans = 0;
	for(int i = 1; i <= n; i++) ans = std::max(ans, dp[h][i]);
	cout << ans;	
	return (0 - 0);
}

P1833 樱花

多重背包板子。

二进制分组优化一下就行了，然后如果说可以无限选的话，就设 k 为一个较大的数就行了。

 /**
 *	author: TLE_Automation
 *	creater: 2022.10.14
 **/
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<bitset>
#include<cstring>
#include<iostream>
#include<algorithm>
#define gc getchar
using namespace std;
typedef long long ll;
const int N = 1e6 + 10;
const int mod = 998244353;
const ll inf = 0x3f3f3f3f3f3f3f3f;
#define debug cout << "i ak ioi" << "\n"
inline void print(int x) {if (x < 0) putchar('-'), x = -x; if(x > 9) print(x / 10); putchar(x % 10 + '0');}
inline char readchar() {static char buf[100000], *p1 = buf, *p2 = buf; return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;}
inline int read() { int res = 0, f = 0; char ch = gc();for (; !isdigit(ch); ch = gc()) f |= (ch == '-'); for (; isdigit(ch); ch = gc()) res = (res << 1) + (res << 3) + (ch ^ '0'); return f ? -res : res;}
 
int n, V, val[N], tim[N], idx = 0, dp[N];
int h1, h2, m1, m2;
 
signed main()
{
	cin >> h1; gc(); cin >> m1; cin >> h2; gc(); cin >> m2;
	if(m2 >= m1) V = (h2 - h1) * 60 + (m2 - m1);
	else h2--, m2 += 60, V = (h2 - h1) *60 + (m2 - m1);
	n = read();
	for(int i = 1; i <= n; i++) {
		int vv = read(), ww = read(), k = read();
		if(!k) k = 1e6;
		for(int j = 1; j <= k; j <<= 1) {
			val[++idx] = ww * j, tim[idx] = vv * j;
			k -= j;
		}
		val[++idx] = ww * k, tim[idx] = vv * k;
	}
	for(int i = 1; i <= idx; i++) {
		for(int j = V; j >= tim[i]; j--) {
			dp[j] = max(dp[j], dp[j - tim[i]] + val[i]);
		}
	}
	cout << dp[V] << "\n";
	return (0 - 0);
}

P1064 [NOIP2006 提高组] 金明的预算方案

有依赖问题的背包dp。

分类讨论。

不买
只卖主件
主件+第一个附件
主件+第二个附件
主件+第一个附件+第二个附件

转移的时候还是 $01$ 背包转移就行了。

 /**
 *	author: TLE_Automation
 *	creater: 2022.10.14
 **/
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<bitset>
#include<cstring>
#include<iostream>
#include<algorithm>
#define mp make_pair
#define pb push_back
#define gc getchar
#define int long long
using namespace std;
typedef long long ll;
const int N = 1e6 + 10;
const int mod = 998244353;
const ll inf = 0x3f3f3f3f3f3f3f3f;
#define debug cout << "i ak ioi" << "\n"
inline void print(int x) {if (x < 0) putchar('-'), x = -x; if(x > 9) print(x / 10); putchar(x % 10 + '0');}
inline char readchar() {static char buf[100000], *p1 = buf, *p2 = buf; return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;}
inline int read() { int res = 0, f = 0; char ch = gc();for (; !isdigit(ch); ch = gc()) f |= (ch == '-'); for (; isdigit(ch); ch = gc()) res = (res << 1) + (res << 3) + (ch ^ '0'); return f ? -res : res;}
 
int V, n, dp[N];
int cnt[N];
pair<int, int> a[N][5]; // first->tim second->val
 /*
 1.不买
 2.只卖主件
 3.主件+1
 4.主件+2
 5.主件+1,2
 */
#define Val(x,y) a[x][y].second
#define Tim(x,y) a[x][y].first
  
signed main()
{
	V = read(), n = read();
	for(int i = 1; i <= n; i++) {
		int tim = read(), val = read(), op = read();
		if(!op) a[i][0].second = val * tim, a[i][0].first = tim;
		else a[op][++cnt[op]].second = val * tim, a[op][cnt[op]].first = tim;
	}	
	for(int i = 1; i <= n; i++) {
		for(int j = V; j >= a[i][0].first; j--) {
			if(j >= Tim(i, 0)) dp[j] = max(dp[j], dp[j - Tim(i, 0)] + Val(i, 0));
			if(j >= Tim(i, 0) + Tim(i, 1)) dp[j] = max(dp[j], dp[j - Tim(i, 0) - Tim(i, 1)] + Val(i, 0) + Val(i, 1));
			if(j >= Tim(i, 0) + Tim(i, 2)) dp[j] = max(dp[j], dp[j - Tim(i, 0) - Tim(i, 2)] + Val(i, 0) + Val(i, 2));
			if(j >= Tim(i, 0) + Tim(i, 1) + Tim(i, 2)) dp[j] = max(dp[j], dp[j - Tim(i, 0) - Tim(i, 1) - Tim(i, 2)] + Val(i, 0) + Val(i, 1) + Val(i, 2));
		}				
	}
	cout << dp[V];
	return (0 - 0);
}

P2340 [USACO03FALL]Cow Exhibition G

感觉是道比较精妙的分类讨论背包题？

我们将智商看做是体积，情商看做价值。

设 $dp_{i,j}$ 表示到了第 $i$ 头牛，智商为 $j$ 的最大情商数。

转移方程是 $dp_{i,j} = \max\{dp_{i-1,j}, dp_{j-Tim_i}+Val_i\}$ 。

这样做是会 MLE 的，我们考虑压掉一维，但是这题是有负数的，所以要将整个 dp 数组右移可能的最小智商和。

然后分类讨论来压掉一维：

当 $Tim_i \ge 0$ 时， $dp_{i,j} = \max\{dp_{i-1,j}, dp_{j-Tim_i}+Val_i\}$ ，我们压掉一维是想让转移 $dp_j$ 的时候，

$dp_j$ 和 $dp_{j-Tim_i}$ 都未被转移过，所以倒叙枚举。

当 $Tim_i < 0$ 时，正序枚举能保证用到的状态都是未被转移过的，也就是上一层的状态。

 /**
 *	author: TLE_Automation
 *	creater: 2022.10.19
 **/
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<bitset>
#include<cstring>
#include<iostream>
#include<algorithm>
#define gc getchar
#define Tim(x) a[x].first
#define Val(x) a[x].second
using namespace std;
typedef long long ll;
const int qwq = 4e5;
const int N = 1e6 + 10;
const int mod = 998244353;
const ll inf = 0x3f3f3f3f3f3f3f3f;
#define debug cout << "i ak ioi" << "\n"
inline void print(int x) {if (x < 0) putchar('-'), x = -x; if(x > 9) print(x / 10); putchar(x % 10 + '0');}
inline char readchar() {static char buf[100000], *p1 = buf, *p2 = buf; return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;}
inline int read() { int res = 0, f = 0; char ch = gc();for (; !isdigit(ch); ch = gc()) f |= (ch == '-'); for (; isdigit(ch); ch = gc()) res = (res << 1) + (res << 3) + (ch ^ '0'); return f ? -res : res;}
 
 
/*
	因为有负数，考虑将整个数组右移 1000*400
	dp_j = dp_{j - v_i} + w_i
	dp_j 表示智商为 j 的最大情商
*/
int n, dp[N << 1];
pair <int, int> a[N];
 
signed main()
{
	n = read();
	memset(dp, -0x3f, sizeof dp);
	dp[qwq] = 0;
	for(int i = 1; i <= n; i++) a[i].first = read(), a[i].second = read();
	for(int i = 1; i <= n; i++) {
		if(Tim(i) >= 0) {
			for(int j = qwq << 1; j >= Tim(i); j--) 
				dp[j] = max(dp[j], dp[j - Tim(i)] + Val(i));
		}
		else {
			for(int j = 0; j <= (qwq << 1) - Tim(i) ; j++) 
				dp[j] = max(dp[j], dp[j - Tim(i)] + Val(i));
		}
	}
	int ans = -0x3f3f3f3f;
	for(int i = qwq; i <= qwq << 1; i++) {
		if(dp[i] >= 0) ans = max(dp[i] + i - qwq, ans);
	}
	cout << ans;
	return (0 - 0);
}

P4059 [Code+#1]找爸爸

因为 $g(k)$ 的性质可以看成是：

如果是第一个空格，就 $-A$ 。
不是第一个空格，就 $-B$ 。

我们设 $dp[i][j][0/1/2]$ 分别表示匹配到了上面串的第 $i$ 个位置，下面串的第 $j$ 个位置，两个都没有空格，上面的有空格，下面的有空格。

转移的话就从对应的状态转移来。

注意一开始赋值极小值，因为空格的贡献是负的。

 #include <bits/stdc++.h>
typedef long long ll;
using namespace std;
constexpr int Maxn = 3e3 + 10;
constexpr ll inf = 0x3f3f3f3f3f3f3f3f;
 
ll dp[Maxn][Maxn][3];
char s[Maxn], t[Maxn];
int val[5][5], n, m, A, B;
 
inline int getpos(char s) {
	if (s == 'A') return 1;
	else if (s == 'T') return 2;
	else if (s == 'G') return 3;u
	else if (s == 'C') return 4;
	else return 114514;
}
 
int main() {
	scanf("%s %s", s + 1, t + 1);
	n = strlen(s + 1), m = strlen(t + 1);
	for (int i = 1; i <= 4; i++) for (int j = 1; j <= 4; j++) scanf("%d", &val[i][j]);
	scanf("%d %d", &A, &B);
	memset(dp, -0x3f, sizeof dp);
	dp[0][0][0] = 0;
	for (int i = 0; i <= n; i++) {
		for (int j = 0; j <= m; j++) {
			if(i && j) dp[i][j][0] = std::max(dp[i - 1][j - 1][0], std::max(dp[i - 1][j - 1][1], dp[i - 1][j - 1][2])) + val[getpos(s[i])][getpos(t[j])];
			if(j) dp[i][j][1] = std::max(dp[i][j - 1][0] - A, std::max(dp[i][j - 1][1] - B, dp[i][j - 1][2] - A));
			if(i) dp[i][j][2] = std::max(dp[i - 1][j][0] - A, std::max(dp[i - 1][j][1] - A, dp[i - 1][j][2] - B));			
		}
	}
	cout << std::max(dp[n][m][0], std::max(dp[n][m][1], dp[n][m][2]));
	return 0;	
}

P3146 [USACO16OPEN]248 G

设 $dp_{l,r}$ 为合并区间 $[l, r]$ 的最大值。

 #include<bits/stdc++.h>
using namespace std;
int dp[500][500], n, a[500];
int main() {
	cin >> n;
	for(int i = 1; i <= n; i++) cin >> a[i];
	memset(dp, -0x3f, sizeof dp);
	for(int i = 1; i <= n; i++) dp[i][i] = a[i];
	for(int len = 2; len <= n; len++) {
		for(int l = 1, r; l + len - 1 <= n; l++) {
			r = l + len - 1;
			for(int k = l; k <= r; k++)
				if(dp[l][k] == dp[k + 1][r]) dp[l][r] = max(dp[l][r], dp[l][k] + 1);
		}
	}
	int ans = -0x3f3f3f3f;
	for(int l = 1; l <= n; l++) for(int r = 1; r <= n; r++) ans = max(ans, dp[l][r]);
	cout << ans; return 0;	 
}

P3147 [USACO16OPEN]262144 P

上一题的加强版，设 $dp_{i,j}$ 表示以 $j$ 为左端点合并出 $i$ 的右端点的右侧。

d p_{i, j} = d p_{i - 1, d p_{i - 1, j}}

$dp_{i,j} = dp_{{i-1},{dp_{i-1,j}}}$

 #include<cstdio>
int dp[60][300010], n, ans;
int main() {
	scanf("%d", &n);
	for(int i = 1, x; i <= n; i++) scanf("%d", &x), dp[x][i] = i + 1;
	for(int i = 2; i <= 58; i++) 
		for(int j = 1; j <= n; j++) {
			if(!dp[i][j]) dp[i][j] = dp[i - 1][dp[i - 1][j]];
			if(dp[i][j]) ans = i;
		}
	printf("%d", ans);	
}

上一篇平衡树学习笔记

下一篇数学做题记录

本文作者：TLE_Automation

本文链接：https://www.cnblogs.com/tttttttle/p/16602154.html

posted @ 2022-08-19 15:29 TLE_Automation 阅读(36) 评论(1) 编辑收藏举报

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TLE_Automationの狗窝

dp 百题计划（11 / 100）

P3205 [HNOI2010]合唱队

P4170 [CQOI2007]涂色

P3842 [TJOI2007]线段

ABC 267 D

P1107 [BJWC2008]雷涛的小猫

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	int n, a[N], dp[1020][1020][2];
	signed main()
	{
	n = read();
	for(int i = 1; i <= n; i++) a[i] = read();
	for(int i = 1; i <= n; i++) dp[i][i][0] = 1;
	for(int len = 2; len <= n; len++) {
	for(int l = 1; l <= n - len + 1; l++) {
	int r = l + len - 1;
	if(a[l] < a[l + 1]) dp[l][r][0] = (dp[l][r][0] + dp[l + 1][r][0]) % mod;
	if(a[l] < a[r]) dp[l][r][0] = (dp[l][r][0] + dp[l + 1][r][1]) % mod;
	if(a[r] > a[r - 1]) dp[l][r][1] = (dp[l][r][1] + dp[l][r - 1][1]) % mod;
	if(a[r] > a[l]) dp[l][r][1] = (dp[l][r][1] + dp[l][r - 1][0]) % mod;
	}
	}
	printf("%d\n", (dp[1][n][1] + dp[1][n][0]) % mod);
	return 0;
	}

	char s[N];
	int n, dp[100][100];

	signed main()
	{
	scanf("%s", s + 1);
	n = strlen(s + 1);
	memset(dp, 0x3f, sizeof dp);
	for(int i = 1; i <= n; i++) dp[i][i] = 1;
	for(int len = 2; len <= n; len++) {
	for(int l = 1, r = l + len - 1; l <= n - len + 1; l++, r++) {
	if(s[l] == s[r]) dp[l][r] = min(dp[l][r - 1], dp[l + 1][r]);
	else
	for(int k = l; k < r; k++) dp[l][r] = min(dp[l][r], dp[l][k] + dp[k + 1][r]);
	}
	}
	printf("%d\n", dp[1][n]);
	return 0;
	}

	#include <bits/stdc++.h>
	using namespace std;
	const int MAXN = 2e4 + 10;

	inline int read() {
	int s = 0, f = 0; char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) f \|= (ch == '-');
	for(; isdigit(ch); ch = getchar()) s = (s << 1) + (s << 3) + (ch ^ '0');
	return f ? -s : s;
	}

	int n, dp[MAXN][2], l[MAXN], r[MAXN];

	int main() {
	n = read();
	for(int i = 1; i <= n; i++) l[i] = read(), r[i] = read();
	dp[1][0] = r[1] + (r[1] - l[1] + 1), dp[1][1] = r[1] - 1;
	for(int i = 2; i <= n; i++) {
	dp[i][0] = min(dp[i - 1][0] + abs(r[i] - l[i - 1]) + abs(r[i] - l[i] + 1), dp[i - 1][1] + abs(r[i] - r[i - 1]) + abs(r[i] - l[i] + 1));
	dp[i][1] = min(dp[i - 1][0] + abs(l[i] - l[i - 1]) + abs(r[i] - l[i] + 1), dp[i - 1][1] + abs(r[i - 1] - l[i]) + abs(r[i] - l[i] + 1));
	}
	return printf("%d\n", min(dp[n][0] + abs(n - l[n]), dp[n][1] + abs(n - r[n]))), 0;
	}

	int n, a[N], m;
	int dp[2020][2020];
	signed main() {

	n = read(), m = read();
	for (int i = 0; i <= n; i++) for (int j = 1; j <= n; j++) dp[i][j] = -inf;
	for (int i = 1; i <= n; i++) a[i] = read();
	for (int i = 1; i <= n; i++) {
	for (int j = 1; j <= min(i, m); j++) {
	dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - 1] + a[i] * j);
	}
	}
	int Max = -inf;
	for (int i = 1; i <= n; i++) Max = max(Max, dp[i][m]);
	cout << Max;
	return (0 - 0);
	}

	/**
	* author: TLE_Automation
	* creater: 2022.10.14
	**/
	#include<cmath>
	#include<queue>
	#include<cstdio>
	#include<vector>
	#include<bitset>
	#include<cstring>
	#include<iostream>
	#include<algorithm>
	#define gc getchar
	using namespace std;
	typedef long long ll;
	const int N = 1e6 + 10;
	const int Maxn = 2e3 + 10;
	const int mod = 998244353;
	const ll inf = 0x3f3f3f3f3f3f3f3f;
	#define debug cout << "i ak ioi" << "\n"
	inline void print(int x) {if (x < 0) putchar('-'), x = -x; if(x > 9) print(x / 10); putchar(x % 10 + '0');}
	inline char readchar() {static char buf[100000], p1 = buf, p2 = buf; return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;}
	inline int read() { int res = 0, f = 0; char ch = gc();for (; !isdigit(ch); ch = gc()) f \|= (ch == '-'); for (; isdigit(ch); ch = gc()) res = (res << 1) + (res << 3) + (ch ^ '0'); return f ? -res : res;}

	int n, h, del, a[Maxn][Maxn], dp[Maxn][Maxn], Max[Maxn];

	/*
	dp[i][j] 表示到了第 i 层高度，第 j 颗树的最大柿子数量
	*/

	signed main()
	{
	n = read(), h = read(), del = read();
	for(int i = 1; i <= n; i++) {
	int x = read();
	for(int j = 1; j <= x; j++) a[read()][i]++;
	}
	for(int i = 1; i <= h; i++) {
	for(int j = 1; j <= n; j++) {
	if(i > del) dp[i][j] = std::max(dp[i - 1][j], Max[i - del]) + a[i][j];
	else dp[i][j] = dp[i - 1][j] + a[i][j];
	Max[i] = std::max(Max[i], dp[i][j]);
	}
	}
	int ans = 0;
	for(int i = 1; i <= n; i++) ans = std::max(ans, dp[h][i]);
	cout << ans;
	return (0 - 0);
	}

	/**
	* author: TLE_Automation
	* creater: 2022.10.19
	**/
	#include<cmath>
	#include<queue>
	#include<cstdio>
	#include<vector>
	#include<bitset>
	#include<cstring>
	#include<iostream>
	#include<algorithm>
	#define gc getchar
	#define Tim(x) a[x].first
	#define Val(x) a[x].second
	using namespace std;
	typedef long long ll;
	const int qwq = 4e5;
	const int N = 1e6 + 10;
	const int mod = 998244353;
	const ll inf = 0x3f3f3f3f3f3f3f3f;
	#define debug cout << "i ak ioi" << "\n"
	inline void print(int x) {if (x < 0) putchar('-'), x = -x; if(x > 9) print(x / 10); putchar(x % 10 + '0');}
	inline char readchar() {static char buf[100000], p1 = buf, p2 = buf; return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;}
	inline int read() { int res = 0, f = 0; char ch = gc();for (; !isdigit(ch); ch = gc()) f \|= (ch == '-'); for (; isdigit(ch); ch = gc()) res = (res << 1) + (res << 3) + (ch ^ '0'); return f ? -res : res;}


	/*
	因为有负数，考虑将整个数组右移 1000*400
	dp_j = dp_{j - v_i} + w_i
	dp_j 表示智商为 j 的最大情商
	*/
	int n, dp[N << 1];
	pair <int, int> a[N];

	signed main()
	{
	n = read();
	memset(dp, -0x3f, sizeof dp);
	dp[qwq] = 0;
	for(int i = 1; i <= n; i++) a[i].first = read(), a[i].second = read();
	for(int i = 1; i <= n; i++) {
	if(Tim(i) >= 0) {
	for(int j = qwq << 1; j >= Tim(i); j--)
	dp[j] = max(dp[j], dp[j - Tim(i)] + Val(i));
	}
	else {
	for(int j = 0; j <= (qwq << 1) - Tim(i) ; j++)
	dp[j] = max(dp[j], dp[j - Tim(i)] + Val(i));
	}
	}
	int ans = -0x3f3f3f3f;
	for(int i = qwq; i <= qwq << 1; i++) {
	if(dp[i] >= 0) ans = max(dp[i] + i - qwq, ans);
	}
	cout << ans;
	return (0 - 0);
	}

	#include<bits/stdc++.h>
	using namespace std;
	int dp[500][500], n, a[500];
	int main() {
	cin >> n;
	for(int i = 1; i <= n; i++) cin >> a[i];
	memset(dp, -0x3f, sizeof dp);
	for(int i = 1; i <= n; i++) dp[i][i] = a[i];
	for(int len = 2; len <= n; len++) {
	for(int l = 1, r; l + len - 1 <= n; l++) {
	r = l + len - 1;
	for(int k = l; k <= r; k++)
	if(dp[l][k] == dp[k + 1][r]) dp[l][r] = max(dp[l][r], dp[l][k] + 1);
	}
	}
	int ans = -0x3f3f3f3f;
	for(int l = 1; l <= n; l++) for(int r = 1; r <= n; r++) ans = max(ans, dp[l][r]);
	cout << ans; return 0;
	}

	#include<cstdio>
	int dp[60][300010], n, ans;
	int main() {
	scanf("%d", &n);
	for(int i = 1, x; i <= n; i++) scanf("%d", &x), dp[x][i] = i + 1;
	for(int i = 2; i <= 58; i++)
	for(int j = 1; j <= n; j++) {
	if(!dp[i][j]) dp[i][j] = dp[i - 1][dp[i - 1][j]];
	if(dp[i][j]) ans = i;
	}
	printf("%d", ans);
	}

TLE_Automationの 狗窝

dp 百题计划 （11 / 100）

P3205 [HNOI2010]合唱队

P4170 [CQOI2007]涂色

P3842 [TJOI2007]线段

ABC 267 D

P1107 [BJWC2008]雷涛的小猫

P1833 樱花

P1064 [NOIP2006 提高组] 金明的预算方案

P2340 [USACO03FALL]Cow Exhibition G

P4059 [Code+#1]找爸爸

P3146 [USACO16OPEN]248 G

P3147 [USACO16OPEN]262144 P

公告

常用链接

随笔档案 (73)

阅读排行榜

评论排行榜

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最新评论

TLE_Automationの狗窝

dp 百题计划（11 / 100）