PTA-L2-007 家庭房产 (25分)
这道题，简单并查集。就是输出处理麻烦了些许。
我们把个人的id作为数组的下标。然后进行存储，合并好之后，用set容器把每个家庭的代表的id存起来，然后相应的存入到结构体中，然后sort一下即可。
代码部分：
#include <bits/stdc++.h>
#define mst(a, n) memset(a, n, sizeof(a))
using namespace std;
const int N = 1e4+ 10;
const int M = 55;
const int INF = 1e6 + 10;
typedef long long ll;
struct node
{
int num;
int peo;
double sumh;
double sumare;
}a[N];
int n;
int id, fath, moth, k, chi;
int fa[N];
int vis[N];
int cnt[N];
int area[N];
int sum[N];
int find(int x)
{
if (x != fa[x])
{
return fa[x] = find(fa[x]);
}
return fa[x];
}
void merge(int x, int y)
{
int fax = find(x);
int fay = find(y);
if (fax > fay)
{
fa[fax] = fay;
}
else
{
fa[fay] = fax;
}
}
bool cmp(node x, node y)
{
if (x.sumare == y.sumare)
{
return x.num < y.num;
}
return x.sumare > y.sumare;
}
int main()
{
for (int i = 0; i < N; i++)
{
fa[i] = i;
}
cin >> n;
for (int i = 0; i < n; i++)
{
cin >> id >> fath >> moth >> k;
vis[id] = 1;
if (fath != -1)
{
vis[fath] = 1;
merge(id, fath);
}
if (moth != -1)
{
vis[moth] = 1;
merge(id, moth);
}
for (int j = 0; j < k; j++)
{
int child;
cin >> child;
vis[child] = 1;
merge(id, child);
}
cin >> cnt[id] >> area[id];
}
set<int> st;
for (int i = 0; i < N; i++)
{
if (!vis[i])
{
continue;
}
int root = find(i);
sum[root]++;
if (root != i)
{
cnt[root] += cnt[i];
area[root] += area[i];
}
st.insert(root);
}
int count = 0;
set<int>::iterator it = st.begin();
while (it != st.end())
{
int root = *it;
a[count].num = root;
a[count].peo = sum[root];
a[count].sumare = 1.0 * area[root] / sum[root];
a[count++].sumh = 1.0 * cnt[root] / sum[root];
it++;
}
sort(a, a + count, cmp);
cout << count << endl;
for (int i = 0; i < count; i++)
{
printf ("%04d %d %.3f %.3f\n", a[i].num, a[i].peo, a[i].sumh, a[i].sumare);
}
return 0;
}