那么我们可以 gS=(−1)|S|−1[S is independent set] 作为容斥系数进行集合幂级数卷积,即: fS=∑T⊆SgTfS∖T 这题中允许加入 ci=0 的边,那么我们依然枚举 S,假如 S 的导出子图中有边,显然必须用等号缩成一个新的点,设 V(S) 为导出子图的连通块数,那么容斥系数 gS=(−1)V(S)−1。 剩下的问
if(m[num][j]==1&&((1<<j)^i)!=i)可以配对但还没转移=>f[(1<<j)^i]=(f[(1<<j)^i]+f[i])%Mod; P - Independent Set(简单树形dp) f[i][0]和f[i][1]存方案 子节点方案乘起来 Q - Flowers(上升子序列附加值) 对于每个i有高度h[i]和值a[i],设f[i]表示1-i的上升子序列最大...
dp P - Independent Set DP dp Q - Flowers DP, Segment Tree dp R - Walk DP dp S - Digit Sum 桁DP dp T - Permutation DP dp W - Intervals DP, Lazy Segment Tree dp X - Tower DP dp Y - Grid 2 DP, 包除原理 dp Z - Frog 3 DP, Convex Hull Trick colopl2018-final-open C - ...
P - Independent Set Q - Flowers Binary Indexed TreeかSegment Treeを使う R - Walk 頂点数が少ない(〜50)ので行列累乗 S - Digit Sum 桁DP T - Permutation 桁DPの類似? U - Grouping ビットDP V - Subtree 全方位木DP W - Intervals ...
This can be done by follows for every b , any Failing Prefix of b-1 will have z=2*(b-1) + K balls , now while adding one more B and W ball to it can B added to any of the z+1 positions and then W can be added to any of the z+2 positions followed by substraction numbe...
Also, it is not hard to accidentally write an easy problem that someone has come up with before. I never said thatkort0ncopied the problem from the Div. 3 round. Obviously it's just coincidence that the exact same problem had occured before. Yet in the future, don't you think some ...
P - Independent Set 树形DP 题意 给定一颗树,可以把树上的每一个节点染色为黑色或白色,但不允许两个相邻的节点同时为黑色。 询问染色方案数。 解法 挺基础的树形DP, 令\(f1[i]\) 表示 \(i\) 为白色时以 \(i\) 为根节点的子树的染色方案。 令\(f2[i]\) 表示 \(i\) 为黑色时以 \(i\) 为...
P.Independent Set类似没有上司的舞会,树形dp。#include <bits/stdc++.h> using namespace std; using ll = long long; constexpr int mod = 1e9 + 7; ll dp[100010][2]; void solve() { int n; cin >> n; vector<vector<int>> g(n + 1); for (int i = 0; i < n - 1;i++) {...
P: Independent Set 思路 很典的树形 DP。 设fu,0/1fu,0/1 表示节点 uu 颜色为 0/10/1 时的方案数(00 白11 黑)。 转移方程: fu,0=∑v is u′s sonfv,0+fv,1fu,1=∑v is u′s sonfv,0fu,0=∑v is u′s sonfv,0+fv,1fu,1=∑v is u′s sonfv,0 代码 点击查看代码 const ...
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