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文章目录
第 42 场周赛
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A AcWing 4311. 最小值
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思路
模拟
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代码
''' Author: NEFU AB-IN Date: 2022-03-12 18:59:53 FilePath: \ACM\Acwing\42\b.py LastEditTime: 2022-03-12 19:02:02 ''' n, m = map(int, input().split()) minn = int(2e9) for i in range(n): a, b = map(int, input().split()) minn = min(minn, a / b) print(f'{m * minn:.6f}')
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B AcWing 4312. 出现次数
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思路
思路是直接KMP,复杂度大约1e8可以过,但不保险
正解是 O ( n 2 ) O(n^2) O(n2)处理出来模板串的前缀和,之后在进行 O ( 1 ) O(1) O(1)的查询 -
代码
''' Author: NEFU AB-IN Date: 2022-03-12 18:59:53 FilePath: \ACM\Acwing\42\b.py LastEditTime: 2022-03-12 19:13:46 ''' n, m, q = map(int, input().split()) s = " " + input() t = " " + input() N = 2000 ne = [0] * N j = 0 for i in range(2, m + 1): while j and t[i] != t[j + 1]: j = ne[j] if t[i] == t[j + 1]: j += 1 ne[i] = j for _ in range(q): l, r = map(int, input().split()) j, ans = 0, 0 for i in range(l, r + 1): while j and s[i] != t[j + 1]: j = ne[j] if s[i] == t[j + 1]: j += 1 if j == m: ans += 1 j = ne[j] print(ans)
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C AcWing 4313. 满二叉树等长路径
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思路
没有推结论,看数据比较小,直接DFS爆搜的
- 首先,先求出每个点上面路径的权值和,这样可以顺便求出来根节点到子节点的最大值 maxn
- 其次,求出每个点为根节点的子节点的最大值,不包括根节点
- 最后从根节点爆搜即可,因为从根节点开始加边的权值是最优的
- 其次每次加边不能让子节点的最大值超过maxn,所以记录差值为 m a x n − ( w [ u ] + w + m a x w [ v ] ) maxn - (w[u] + w + maxw[v]) maxn−(w[u]+w+maxw[v])
- 即 最大值 - (此节点u上面的权值和 + (u->v)这条边的权值 + 此节点u下面的最大值)
- 比如下面这个例子,当u = 1,v = 2时
cha = 22 - 0 - 1 - 14 = 7
- 之后将 上面补的差值 和 自己补的差值 往下传即可
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代码
''' Author: NEFU AB-IN Date: 2022-03-12 19:29:40 FilePath: \ACM\Acwing\42\c.py LastEditTime: 2022-03-12 21:44:35 ''' N = int(1e4 + 10) g = [[] * N for _ in range(N)] ww = [0] * N # 记录当前点上面的权值 maxw = [0] * N #记录当前点下面的最大值 maxn = 0 maxjd = 1 ans = 0 def dfs1(fa, u, len): global maxn, maxjd for v, w in g[u]: if v == fa: continue ww[v] = ww[u] + w dfs1(u, v, len + w) if len > maxn: maxn = len maxjd = u def dfs2(fa, u): for v, w in g[u]: if v == fa: continue maxw[u] = max(maxw[u], dfs2(u, v) + w) return maxw[u] def dfs(fa, u, z): global ans for v, w in g[u]: if v == fa: continue cha = maxn - ww[u] - maxw[v] - w ww[v] += (cha + z) #差值 和 补充父节点补的 ans += cha dfs(u, v, cha + z) n = int(input()) a = list(map(int, input().split())) for i in range(len(a)): g[i + 2].append([(i + 2) // 2, a[i]]) g[(i + 2) // 2].append([i + 2, a[i]]) dfs1(0, 1, 0) dfs2(0, 1) dfs(0, 1, 0) print(ans)
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