P5007 DDOSvoid 的疑惑

合集 - 图论(79)

1.F. Chat Screenshots2024-02-192.P1656 炸铁路2024-02-223.P1137 旅行计划2024-02-244.P2835 刻录光盘2024-02-245.P1197 [JSOI2008] 星球大战2024-02-246.P3388 【模板】割点（割顶）2024-02-247.P8435 【模板】点双连通分量2024-02-258.P8436 【模板】边双连通分量2024-02-259.P2860 [USACO06JAN] Redundant Paths G2024-02-2510.P1653 [USACO04DEC] Cow Ski Area G2024-02-2611.P3047 [USACO12FEB] Nearby Cows G2024-03-0412.P1894 [USACO4.2] 完美的牛栏The Perfect Stall2024-03-0613.P1550 [USACO08OCT] Watering Hole G2024-03-0614.P2330 [SCOI2005] 繁忙的都市2024-03-0615.P1525 [NOIP2010 提高组] 关押罪犯2024-03-0716.P1379 八数码难题2024-03-0717.P2746 [USACO5.3] 校园网Network of Schools2024-03-0818.P6121 [USACO16OPEN] Closing the Farm G2024-03-0819.P2341 [USACO03FALL / HAOI2006] 受欢迎的牛 G2024-03-0820.P2055 [ZJOI2009] 假期的宿舍2024-03-0821.P5905 【模板】全源最短路（Johnson）2024-03-1122.F. Microcycle2024-03-1223.G. Path Prefixes2024-03-1424.G. Rudolf and Subway2024-03-1525.C. Ehab and Path-etic MEXs2024-03-1726.A. String Transformation 12024-03-1727.D. Secret Passwords2024-03-1928.F. Maximum White Subtree2024-03-1929.P3478 [POI2008] STA-Station2024-03-1930.P1347 排序2024-03-2131.P1960 郁闷的记者2024-03-2132.E1. Weights Division (easy version)2024-03-22

33.P5007 DDOSvoid 的疑惑2024-03-25

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收起

原题链接

题解

1.具体去考虑每个集合所包含的元素及其大小个数是非常繁琐的，所以我们考虑每个元素对答案的贡献
2.
令 $f[now]$ 代表以 $now$ 为根节点的答案
$sizes[now]$ 代表以 $now$ 为根节点所包含集合的个数
更新过程如下：
$f[now]+=f[next]\ast (sizes[now]+1)+f[now]\ast sizes[next]$

$size[now]+=sizes[next]\ast (sizes[now]+1)$

最后 $f[now]+=a[now],sizes[next]++$

就像往一颗已知树插入子树的过程

code

#define ll long long
#include<bits/stdc++.h>
using namespace std;
const ll mod=1e8+7;
ll f[1000006]={0},sizes[1000006]={0};

vector<ll> G[1000006];
ll k,n;
void dfs(ll now,ll fa)
{
    for(auto next:G[now])
    {
        if(next==fa) continue;
        dfs(next,now);

        f[now]=(f[now]+f[next]*(sizes[now]+1)%mod)%mod+f[now]*(sizes[next])%mod;
        sizes[now]+=sizes[next]*(sizes[now]+1)%mod;
        f[now]%=mod;
        sizes[now]%=mod;
    }
    f[now]+=(k?now:1);
    f[now]%=mod;
    sizes[now]++;
    sizes[now]%=mod;
}

int main()
{
    ios::sync_with_stdio(false);
    cin>>n>>k;
    for(ll i=1;i<n;i++)
    {
        ll x,y;
        cin>>x>>y;
        G[x].emplace_back(y);
        G[y].emplace_back(x);
    }

    dfs(1,1);
    cout<<f[1];
    return 0;
}