Edgy Trees(dfs,并查集,快速幂,树形结构,红黑树)

合集 - 刷题集(97)
1.妖梦拼木棒2.[USACO1.2]回文平方数 Palindromic Squares3.[COCI2012-2013#1] LJUBOMORA4.火柴棒等式5.Kieszonkowe6.[COCI2014-2015#5] TRAKTOR7.硬币翻转8.[蓝桥杯 2020 省 AB1] 走方格9.P2097 资料分发10.[AHOI2005]约数研究11.[NOIP2004 普及组] 火星人12.我焯,原神(这真是个题)13.滑雪 OJ365114.CV的人有福啦!YTU贪心训练2(部分注释)15.砝码称重16.警钟长鸣-----找零钱17.[USACO09OCT]Barn Echoes G18.【CSGRound2】光骓者的荣耀19.初见蓝桥杯20.[蓝桥杯 2022 省 B] 扫雷21.开餐馆22.P9241 [蓝桥杯 2023 省 B] 飞机降落23.P7083 [NWRRC2013] Energy Tycoon24.高精度25.P7954 [COCI2014-2015#6] PAPRIKA26.(输出路径搜索)[USACO06OCT] Cows on Skates G27.畜栏保留问题28.(贪心+搜索+剪枝)P8801 [蓝桥杯 2022 国 B] 最大数字29.蓝桥杯 入门----普及-30.[NOIP2016 提高组] 玩具谜题31.[NOIP2002 提高组] 字串变换(无向图查重)32.[NOIP2000 提高组] 单词接龙33.POJ 硬币34.动态规划学习 1(最长上升子序列问题)35.动态规划学习 2(背包问题及其变形)36. 切割回文37.[NOIP2001 提高组] 数的划分(剪枝)38.字符串转换数字,sscanf和sprintf大法39.[CTSC1997] 选课(树状DP)40. k 倍区间(同余定理,组合数)41.[蓝桥杯 2018 省 AB] 全球变暖42.马的遍历(对bfs的更深一层探讨)43.找零钱44.Pasha Maximizes45.Newspaper Headline46.Balanced Ternary String47.(数论)素数,质数48.(数论) 约数49.Helping the Nature50.(博弈论)Even Number Addicts51.Present52.Mr. Kitayuta's Colorful Graph(二维并查集,弱化版)
53.Edgy Trees(dfs,并查集,快速幂,树形结构,红黑树)
54.并查集55.再谈 前缀和,差分,离散化56.Best Cow Fences(前缀和+特殊二分)57.二分58.同类型,类背包动态规划,选地dp59.Primes on Interval(欧拉筛+二分+滑动窗口)60.字符串(KMP , Trie树 , STL)61.Codeforces1 #879 div.262.Add Modulo 10(数论,思维,数学,规律)63.期末考试YTU4035: Shmily(数学,等差数列)64.[POI2006] OKR-Periods of Words65.Getting Zero(Bfs)66.快速幂67.Dreaming of Freedom(数论,贪心)68.Scoring Subsequences69.Dolce Vita70.Cake Assembly Line71.Constructive Problem72.Forever Winter73.Maximum Sum(前缀和)74.Contrast Value(数学规律)75.Lamps(STL+双端队列)76.Long Long77.Sum in Binary Tree78.Apple Tree(树状搜索,树形DP)79.Strong Password(贪心思想)80.Come Together81.Notepad82.Hamiltonian Wall83.7.17作业84.逛画展(双指针)85.#88586.树状数组87.Subsequence Addition88.7.22作业89.Diverse Substrings90.Tracking Segments(二分,区间前缀)91.最大食物链计数92.线段树193.[TJOI2007] 线段(状态机)94.#87895.测试196.李白打酒97.砍竹子
收起
Edgy Trees
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given a tree (a connected undirected graph without cycles) of n vertices. Each of the n−1−1 edges of the tree is colored in either black or red.

You are also given an integer k. Consider sequences of k vertices. Let's call a sequence [a1,a2,…,ak][1,2,…,] good if it satisfies the following criterion:

  • We will walk a path (possibly visiting same edge/vertex multiple times) on the tree, starting from a11 and ending at ak.
  • Start at a11, then go to a22 using the shortest path between a1 and a2, then go to a3 in a similar way, and so on, until you travel the shortest path between ak−1−1 and ak.
  • If you walked over at least one black edge during this process, then the sequence is good.
 

Consider the tree on the picture. If k=3=3 then the following sequences are good: [1,4,7], [5,5,3] and [2,3,7][2,3,7]. The following sequences are not good: [1,4,6][1,4,6], [5,5,5][5,5,5], [3,7,3]

There are nk sequences of vertices, count how many of them are good. Since this number can be quite large, print it modulo 109+7109+7.

Input

The first line contains two integers n and k (2≤n≤1052≤≤105, 2≤k≤1002≤≤100), the size of the tree and the length of the vertex sequence.

Each of the next n−1lines contains three integers ui, vi and xi (1≤ui,vi≤n1≤,≤, xi∈{0,1}∈{0,1}), where ui and vi denote the endpoints of the corresponding edge and xi is the color of this edge (00 denotes red edge and 11 denotes black edge).

Output

Print the number of good sequences modulo 109+7109+7.

Examples
input Copy
4 4
1 2 1
2 3 1
3 4 1
output Copy
252
input Copy
4 6
1 2 0
1 3 0
1 4 0
output Copy
0
input Copy
3 5
1 2 1
2 3 0
output Copy
210
Note

In the first example, all sequences (4444) of length 44 except the following are good:

  • [1,1,1,1]
  • [2,2,2,2]
  • [3,3,3,3]
  • [4,4,4,4]
复制代码
//https://www.luogu.com.cn/problem/CF1139C
//dfs解法
//用dfs记录下来一个连通块的节点数,然后用幂运算算出不合法的方案数
#include<bits/stdc++.h>
#define ll long long 
using namespace std;
const int N=1e5+10,mod=1e9+7;
int n,k,idx;
ll res,num;
vector<int>mp[N];
bool vis[N];
//快速幂
// ll qpow(ll x,ll y){
//     ll ba=x,ans=1;
//     while(y){
//         if(y&1) ans=(ans*ba)%mod;
//         ba=(ba*ba)%mod;
//         y>>=1;
//     }
//     return ans;
// }

//普通幂
ll qpow(int x,int cc)
{
    ll ans=1;
    for(int i=0;i<cc;i++) ans=(ans*x)%mod;
    return ans;
}
void dfs(int x)
{
    vis[x]=true;
    ++idx;
    for(auto i:mp[x]) if(!vis[i]) dfs(i);
}
int main()
{
    std::ios::sync_with_stdio(false);
    cin>>n>>k;
    for(int i=1;i<n;i++){
        int a,b,c;
        cin>>a>>b>>c;
        if(c==0) mp[a].push_back(b), mp[b].push_back(a);
    }
    res=qpow(n%mod,k)%mod;
    for(int i=1;i<=n;i++)
        if(!vis[i]){
            idx=0;
            dfs(i);
            num=(num+qpow(idx,k))%mod;
        }
    cout<<(res-num+mod)%mod;
    return 0;
}
复制代码
复制代码
//并查集解法:
//用并查集来维护是否合法
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=1e5+10,mod=1e9+7;
int p[N],n,k,vis[N],cnt,idx,mp[N];
bool st[N];
ll res,num;
int find(int x)
{
    if(p[x]!=x) p[x]=find(p[x]);
    return p[x];
}
ll qpow(int x)
{
    ll ans=1;
    for(int i=0;i<k;i++) ans=(ans*x)%mod;
    return ans;
}
int main()
{
    std::ios::sync_with_stdio(false);
    cin>>n>>k;
    for(int i=1;i<=n;i++) p[i]=i;
    for(int i=1;i<=n;i++){
        int a,b,c;
        cin>>a>>b>>c;
        if(c==0) p[find(a)]=find(b);//如果是红的就加进去
    }
    res=qpow(n%mod);
    for(int i=1;i<=n;i++) mp[find(i)]++;//新方法注意,如何找出这个点的连通块数量?
    //很明显是这样找得到啦,建立一个桶;
    for(int i=1;i<=n;i++) num=(num+qpow(mp[i]))%mod;
    cout<<(res-num+mod)%mod<<endl;
    return 0;
}
复制代码

 

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本文作者：Sakurajimamai
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