L2-007. 家庭房产

题目链接：https://www.patest.cn/contests/gplt/L2-007

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<set>
#include<algorithm>

using namespace std;
const int maxn=10007;
const int INF=0x3f3f3f3f;
int vis[maxn], p[maxn], m[maxn];
struct node
{
    int id, num;
    double ans1, ans2;
} a[maxn], b[maxn], mul[maxn];

void init()
{
    memset(vis, 0, sizeof(vis));
    memset(mul, 0, sizeof(mul));
    for(int i=0; i<10000; i++)
        p[i]=i;
}

int cmp(node a, node b)
{
    if(a.ans2!=b.ans2)
        return a.ans2>b.ans2;
    return a.id<b.id;
}
int Find(int x)
{
    while(x!=p[x])
        x=p[x];
    return x;
}


void solve(int a, int b)
{
    int x=Find(a);
    int y=Find(b);
    if(x!=y)
    {
        if(x>y)
            p[x]=y;
        else
            p[y]=x;
    }
}
int main()
{
    int n;
    scanf("%d", &n);
    init();
    for(int i=0; i<n; i++)
    {
        int p1, p2, m, d;
        scanf("%d %d %d", &a[i].id, &p1, &p2);
        vis[a[i].id]=1;
        if(p1!=-1)
        {
            solve(a[i].id, p1);
            vis[p1]=1;
        }
        if(p2!=-1)
        {
            solve(a[i].id, p2);
            vis[p2]=1;
        }
        scanf("%d", &m);
        while(m--)
        {
            scanf("%d", &d);
            if(d!=-1)
            {
                solve(a[i].id, d);
                vis[d]=1;
            }
        }
        scanf("%lf %lf", &a[i].ans1, &a[i].ans2);
    }
    for(int i=0; i<n; i++)
    {
        int id=Find(a[i].id);
        mul[id].id=id;
        mul[id].ans1+=a[i].ans1;
        mul[id].ans2+=a[i].ans2;
    }
    for(int i=0; i<10000; i++)
    {
        if(vis[i])
            mul[Find(i)].num++;
    }
    int cnt=0;
    for(int i=0; i<10000; i++)
    {
        if(vis[i])
        {
            int id=Find(i);
            if(!m[id])
            {
                m[id]=1;
                double x=double(mul[id].num);
                b[cnt].id=id;
                b[cnt].ans1=mul[id].ans1*1.0/x;
                b[cnt].ans2=mul[id].ans2*1.0/x;
                b[cnt++].num=mul[id].num;
            }
        }
    }
    sort(b, b+cnt, cmp);
    printf("%d\n", cnt);
    for(int i=0; i<cnt; i++)
        printf("%04d %d %.3f %.3f\n", b[i].id, b[i].num, b[i].ans1, b[i].ans2);
    return 0;
}