Codeforces-Round#548(Div.2)-C-Edgy Trees-快速幂

这个题没想出来，好菜QAQ...

题目要求至少经过一段黑色的边的答案，那么我们可以求总数减去不经过黑色的，这样就是答案了。

那么也就是求出每个只有红色边的连通块中有几个点。

总数就是n的k次方，设连通块有i个点，那么每次减去i的k次方就行了。

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <iomanip>
#include <string>using namespace std;typedef long long int ll;
ll n,k,sum;
int u,v,x;
const int maxn=100000+5;
vector<int> edge[maxn];
const int mod=1e9+7;
int vis[maxn];ll quick_m(ll a,ll b){ll res=1ll;while(b){if(b%2) res=res*a%mod;a=a*a%mod;b/=2;}return res;
}void DFS(int x){vis[x]=1;sum++;for(int i=0;i<edge[x].size();i++){int nxt=edge[x][i];if(!vis[nxt]) DFS(nxt);}
}int main(){while(scanf("%I64d%I64d",&n,&k)!=EOF){for(int i=0;i<=n;i++) edge[i].clear();memset(vis,0,sizeof(vis));for(int i=1;i<n;i++){scanf("%d%d%d",&u,&v,&x);if(x==0){//去掉黑色的边edge[u].push_back(v);edge[v].push_back(u);}}ll ans=quick_m(n,k);for(int i=1;i<=n;i++){if(!vis[i]){sum=0;DFS(i);ans=((ans-quick_m(sum,k))%mod+mod)%mod;}}printf("%I64d\n",ans);}return 0;
}

转载于:https://www.cnblogs.com/JustDoA/p/10586090.html

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