题解 CF833D Red-Black Cobweb
题目大意
给出一个 \(n\) 个点的树,每条边有边权和颜色 \(0,1\) ,定义一条链合法当且仅当 \(0,1\) 颜色的边数之比小于等于 \(2\) ,求所有合法的链的边权之积的积。
\(n\le 10^5\),答案对 \(10^9+7\) 取模。
思路
边分治板题,但是因为边界问题爆炸了。。。
首先先容斥一下,即总答案除以不合法答案,然后你发现总答案特别好求,不合法方案可是使用边分治解决。
时间复杂度 \(\Theta(n\log^2 n)\) 。
\(\texttt{Code}\)
#include <bits/stdc++.h>
using namespace std;
#define Int register int
#define mod 1000000007
#define MAXN 800005
template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
int n,ans = 1;
int qkpow (int a,int b){
int res = 1;for (;b;b >>= 1,a = 1ll * a * a % mod) if (b & 1) res = 1ll * res * a % mod;
return res;
}
int inv (int x){return qkpow (x,mod - 2);}
namespace Graph{
#define PII pair<int,int>
int cnt = 1,toop = 1,pres[MAXN],to[MAXN << 1],wei[MAXN << 1],col[MAXN << 1],nxt[MAXN << 1],head[MAXN],siz[MAXN];bool vis[MAXN];
void Add_Edge (int u,int v,int w,int c){
to[++ toop] = v,wei[toop] = w,col[toop] = c,nxt[toop] = head[u],head[u] = toop;
to[++ toop] = u,wei[toop] = w,col[toop] = c,nxt[toop] = head[v],head[v] = toop;
}
struct node{
int R,B,dis;
};
node *f,T1[MAXN],T2[MAXN];
void dfs (int u,int fa,int totr,int totb,int pre){
if (u <= n) f[++ cnt] = node {totr,totb,pre};
for (Int i = head[u];i;i = nxt[i]){
int v = to[i];
if (v == fa || vis[i]) continue;
dfs (v,u,totr + (col[i] == 0),totb + (col[i] == 1),1ll * pre * wei[i] % mod);
}
}
int ed,lim,Siz,lena,lenb;
void findedge (int u,int fa){//找重边
siz[u] = 1;
for (Int i = head[u];i;i = nxt[i]){
int v = to[i];
if (v == fa || vis[i]) continue;
findedge (v,u),siz[u] += siz[v];
int tmp = max (siz[v],Siz - siz[v]);
if (tmp < lim) ed = i,lim = tmp;
}
}
bool cmp1 (node a,node b){return 2 * a.B - a.R < 2 * b.B - b.R;}
bool cmp2 (node a,node b){return 2 * a.R - a.B < 2 * b.R - b.B;}
void Solve (int u,int S){
if (S <= 1) return ;
lim = Siz = S,findedge (u,0),vis[ed] = vis[ed ^ 1] = 1;
cnt = 0,f = T1,dfs (to[ed],0,0,0,1),lena = cnt;
cnt = 0,f = T2,dfs (to[ed ^ 1],0,0,0,1),lenb = cnt;
for (Int i = 1;i <= lenb;++ i) T2[i].R += (col[ed] == 0),T2[i].B += (col[ed] == 1),T2[i].dis = 1ll * T2[i].dis * wei[ed] % mod;
sort (T1 + 1,T1 + lena + 1,cmp1);
pres[0] = 1;for (Int i = 1;i <= lena;++ i) pres[i] = 1ll * pres[i - 1] * T1[i].dis % mod;
for (Int i = 1;i <= lenb;++ i){
int now = T2[i].R - 2 * T2[i].B,l = 1,r = lena,fuckans = 0;
while (l <= r){
int mid = (l + r) >> 1;
if (2 * T1[mid].B - T1[mid].R < now) fuckans = mid,l = mid + 1;
else r = mid - 1;
}
ans = 1ll * ans * qkpow (T2[i].dis,fuckans) % mod * pres[fuckans] % mod;
}
sort (T1 + 1,T1 + lena + 1,cmp2);
pres[0] = 1;for (Int i = 1;i <= lena;++ i) pres[i] = 1ll * pres[i - 1] * T1[i].dis % mod;
for (Int i = 1;i <= lenb;++ i){
int now = T2[i].B - 2 * T2[i].R,l = 1,r = lena,fuckans = 0;
while (l <= r){
int mid = (l + r) >> 1;
if (2 * T1[mid].R - T1[mid].B < now) fuckans = mid,l = mid + 1;
else r = mid - 1;
}
ans = 1ll * ans * qkpow (T2[i].dis,fuckans) % mod * pres[fuckans] % mod;
}
int tx = to[ed],ty = to[ed ^ 1];
if (siz[tx] > siz[ty]) siz[tx] = S - siz[ty];
else siz[ty] = S - siz[tx];
Solve (tx,siz[tx]),Solve (ty,siz[ty]);
}
}
int cnt,all = 1,toop = 1,to[MAXN << 1],wei[MAXN << 1],col[MAXN << 1],nxt[MAXN << 1],head[MAXN],las[MAXN],siz[MAXN];
void Add_Edge (int u,int v,int w,int c){
to[++ toop] = v,wei[toop] = w,col[toop] = c,nxt[toop] = head[u],head[u] = toop;
to[++ toop] = u,wei[toop] = w,col[toop] = c,nxt[toop] = head[v],head[v] = toop;
}
void dfs (int u,int fa){
siz[u] = 1;
for (Int i = head[u];i;i = nxt[i]){
int v = to[i],w = wei[i],c = col[i];
if (v == fa) continue;
if (!las[u]) las[u] = u,Graph::Add_Edge (u,v,w,c);
else ++ cnt,Graph::Add_Edge (las[u],cnt,1,-1),Graph::Add_Edge (las[u] = cnt,v,w,c);
dfs (v,u),siz[u] += siz[v],all = 1ll * all * qkpow (w,1ll * siz[v] * (n - siz[v]) % (mod - 1)) % mod;
}
}
signed main(){
read (n),cnt = n;
for (Int i = 2,u,v,w,c;i <= n;++ i) read (u,v,w,c),Add_Edge (u,v,w,c);
dfs (1,0),Graph::Solve (1,cnt),write (1ll * all * inv (ans) % mod),putchar ('\n');
return 0;
}
题解 CF833D Red-Black Cobweb的更多相关文章
- 【CF833D】Red-Black Cobweb(点分治)
[CF833D]Red-Black Cobweb(点分治) 题面 CF 有一棵树,每条边有一个颜色(黑白)和一个权值,定义一条路径是好的,当且仅当这条路径上所有边的黑白颜色个数a,b满足2min(a, ...
- 【CF833D】Red-Black Cobweb
[CF833D]Red-Black Cobweb 题面 洛谷 题解 看到这种统计路径的题目当然是淀粉质啦. 考虑转化一下信息设一条路径上有红点\(a\)个,黑点\(b\)个 则\(2min(a,b)\ ...
- Hdoj 1312.Red and Black 题解
Problem Description There is a rectangular room, covered with square tiles. Each tile is colored eit ...
- CF833D Red-Black Cobweb
题面 题解 点分治大火题... 设白边数量为$a$,黑边为$b$,则$2min(a,b)\geq max(a,b)$ 即$2a\geq b\;\&\&2b\geq a$ 考虑点分治时如 ...
- 题解报告:hdu 1312 Red and Black(简单dfs)
Problem Description There is a rectangular room, covered with square tiles. Each tile is colored eit ...
- poj 1979 Red and Black 题解《挑战程序设计竞赛》
地址 http://poj.org/problem?id=1979 Description There is a rectangular room, covered with square tiles ...
- 【CF1425B】 Blue and Red of Our Faculty! 题解
原题链接 简要翻译: 有一个连通图,A和B同时从点1出发,沿不同的路径前进.原本,图上的每一条边都是灰色的.A将经过的边涂成红色,B将经过的边涂成蓝色的.每个回合每个人只能走灰色的边.当某个回合中不存 ...
- CF833D Red-Black Cobweb 点分治、树状数组
传送门 统计所有路径的边权乘积的乘积,不难想到点分治求解. 边权颜色比例在\([\frac{1}{2},2]\)之间,等价于\(2B \geq R , 2R \geq B\)(\(R,B\)表示红色和 ...
- 洛谷 CF399B【Red and Blue Balls】题解
n年没有更博客:我总结出了规律,当学的东西很难得时候都去学习,没有时间写博客,只有 内容对于我这种蒟蒻友好,又让我非常闲的慌时才写博客,这种博客以后也没有价值(也有些是做完一道题有成就感写的) 最近内 ...
随机推荐
- mysql最强
MYSQL 与mysql第一次亲密接触 数据库相关概念 一.数据库的好处 二.数据库的常见概念 ★ 三.数据库存储数据的特点 四.常见的数据库管理系统 MYSQL的介绍 一.MySQL的背景 二.My ...
- Matlab实现BP神经网络预测(附实例数据及代码)
BP神经网络介绍 神经网络是机器学习中一种常见的数学模型,通过构建类似于大脑神经突触联接的结构,来进行信息处理.在应用神经网络的过程中,处理信息的单元一般分为三类:输入单元.输出单元和隐含单元. 顾名 ...
- C# - 习题04_分析代码写出结果i1、i2、c.i、str、c.str
时间:2017-08-23 整理:byzqy 题目:分析如下代码,写出程序输出结果. 文件:Class1.cs using System; namespace Interview3 { class C ...
- Clean Architecture For RazorPage 实现多语言和本地化
最近终于把多语言功能加上了,这次就再发一篇,讲一下在asp.net core环境下如何实现多语言和本地化(Globalization and localization)功能,主要参看:ASP.NET ...
- 实型(浮点型):float、double
实型(浮点型):float.double 实型变量也可以称为浮点型,浮点型变量是用来存储小数数值的.在C语言中,浮点型分为两种:单精度浮点型(float).双精度浮点型(double),但是doubl ...
- Jenkins持续集成接口压测
步骤 自动化压测- jmeter + shell Jenkins与jmeter压测,环境要求 自动压测运行逻辑 Jmeter输出压力测试报告 压测报告与Jenkins集成 Jenkins任务:源码同步 ...
- JS012. 变量存储含class、id等其他属性的标签元素(动态渲染DOM结点)
项目中有一处源码需要用变量存储html标签,包含类名和其他一些属性,再动态地将其渲染到页面上. 看下普通的存储方式: initHtml: function () { var me = this; // ...
- poll?transport=longpoll&connection...烦人的请求c
1.问题描述: 最近使用miniui做了一个后台管理系统,打开浏览器调试时,总发现一堆无关的请求,结构大致是:poll?transport=longpoll&connection.....一直 ...
- FastAPI(6)- get 请求 - 详解 Query
可选参数 上一篇文章讲过查询参数可以不是必传的,可以是可选参数 from fastapi import FastAPI from typing import Optional import uvico ...
- 安卓gradle时报错"ERROR: Plugin with id 'com.android.application' not found."
在build.gradle中更改gradle插件版本号 buildscript { repositories { google() jcenter() } dependencies { //版本号请根 ...