团体程序设计天梯赛-练习集 L2-007. 家庭房产

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团体程序设计天梯赛-练习集
L2-007. 家庭房产
https://www.patest.cn/contests/gplt/L2-007
先进行并查集，在并集时将信息赋值给最小的编号，同时记录所有的第一个编号，之后用set把所有第一个的编号的根放入set，直接输出。

#include<iostream>#include<cstring>#include<cstdio>#include<set>using namespace std;void unite(int x, int y);const int maxn = 10005;int par[maxn];int ran[maxn];struct Size {    int g;    double area;}siz[maxn];struct Ans {    int name;    double g;    double area;    bool operator<(const Ans &t1) const {        if (area != t1.area) {            return (area > t1.area);        }        else if (ran[name] != ran[t1.name]) {            return (ran[name] > ran[t1.name]);        }        else {            return (name < t1.name);        }    }}ans;int k;void update(int a) {    int b;    scanf("%d", &b);    if (b == -1) {        return;    }    else {        unite(a, b);    }}void init(int n) {    for (int i = 0; i <= n; i++) {        par[i] = i;        ran[i] = 1;    }}int find(int x) {    int r = x;    while (r != par[r])        r = par[r];    int i = x, j;    while (par[i] != r)    {        j = par[i];        par[i] = r;        i = j;    }    return r;}void unite(int x, int y) {    x = find(x);    y = find(y);    if (x == y) return;    if (x > y) {        par[x] = y;        ran[y] += ran[x];        siz[y].g += siz[x].g;        siz[y].area += siz[x].area;    }    else {        par[y] = x;        ran[x] += ran[y];        siz[x].g += siz[y].g;        siz[x].area += siz[y].area;    }}set<Ans>s;int main() {    int N;    while (scanf("%d", &N) != EOF) {        if (N == 0) {            continue;        }        if (s.size()) {            s.clear();        }        init(10000);        k = 0;        int a[1024];        memset(siz, 0, sizeof(siz));        for (int i = 0; i < N; i++) {            int z;            scanf("%d", &a[i]);            update(a[i]);            update(a[i]);            scanf("%d", &z);            for (int j = 0; j < z; j++) {                update(a[i]);            }            int m;            double ar;            scanf("%d%lf", &m, &ar);            siz[find(a[i])].g += m;            siz[find(a[i])].area += ar;        }        for (int i = 0; i < N; i++) {            ans.name = find(a[i]);            ans.g = (double)siz[ans.name].g/(double)ran[ans.name];            ans.area = (double)siz[ans.name].area / (double)ran[ans.name];            s.insert(ans);        }        printf("%d\n", s.size());        set<Ans>::iterator it;        for (it = s.begin(); it != s.end(); it++) {            printf("%04d %d %.3lf %.3lf\n", (*it).name, ran[(*it).name], (*it).g, (*it).area);        }    }    return 0;}

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