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  • PAT 天梯赛 L2-007. 家庭房产 【并查集】

    题目链接

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

    思路
    将一个家庭里的所有人都并进去
    然后最后查找的时候 找到所有同一个家庭的人,计算出人数,人均房产套数,人均房产面积 而且 在ID 小于当前 ID 的时候 要更新

    AC代码

    #include <cstdio>
    #include <cstring>
    #include <ctype.h>
    #include <cstdlib>
    #include <cmath>
    #include <climits>
    #include <ctime>
    #include <iostream>
    #include <algorithm>
    #include <deque>
    #include <vector>
    #include <queue>
    #include <string>
    #include <map>
    #include <stack>
    #include <set>
    #include <numeric>
    #include <sstream>
    #include <iomanip>
    #include <limits>
    
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    typedef pair <int, int> pii;
    typedef pair <ll, ll> pll;
    
    const double PI  = 3.14159265358979323846264338327;
    const double E   = 2.718281828459;
    const double eps = 1e-6;
    
    const int INF    = 0x3f3f3f3f;
    const int maxn   = 1e6 + 5;
    const int MOD    = 1e9 + 7;
    
    int pre[10000];
    
    int find(int x)
    {
        while (pre[x] != x)
            x = pre[x];                   
        return x; 
    }
    
    void join(int x, int y)
    {
        int fx = find(x), fy = find(y);
        if (x != fy)
            pre[fx] = fy;
    }   
    
    struct Node
    {
        int id;
        int total;
        int tot;
        double ave, area;
    }temp, q[maxn];
    
    bool comp(Node x, Node y)
    {
        if (fabs(x.area - y.area) < eps)
            return x.id < y.id;
        return x.area > y.area;
    }
    
    int main()
    {
        for (int i = 0; i < 10000; i++)
            pre[i] = i;
        int n;
        cin >> n;
        map <int, Node> m;
        map <int, int> vis; 
        for (int i = 0; i < n; i++)
        {
            int id, a, b, k;
            scanf("%d%d%d%d", &id, &a, &b, &k);
            vis[id] = 1;
            if (a != -1)
            {
                join(id, a);
                vis[a] = 1;
            }   
            if (b != -1)
            {
                join(id, b);
                vis[b] = 1;
            }   
            for (int j = 0; j < k; j++)
            {
                scanf("%d", &a);
                if (a != -1)
                {
                    join(id, a);
                    vis[a] = 1;
                }
            }
            scanf("%d%d", &a, &b);
            m[id].id   = id;
            m[id].tot  = a;
            m[id].area = b;
        }
        map <int, int>::iterator it, iter;
        int count = 0;
        for (it = vis.begin(); it != vis.end(); it++)
        {
            temp.id   = 0;
            temp.tot  = 0;
            temp.tot  = 0;
            temp.ave  = 0.0;
            temp.area = 0.0;
            if (it -> second == 1)
            {
                it -> second = 0;
                int flag   = find(it -> first);
                temp.id    = it -> first;
                temp.tot  += m[it -> first].tot;
                temp.area += m[it -> first].area;
                temp.total = 1;
                for (iter  = it, iter++; iter != vis.end(); iter++)
                {
                    if (iter -> second == 1 && flag == find(iter -> first))
                    {
                        iter -> second = 0;
                        if (m[iter -> first].id == iter -> first)
                        {
                            if (iter -> first < temp.id)
                                temp.id = iter -> first;
                            temp.tot  += m[iter -> first].tot;
                            temp.area += m[iter -> first].area;
                        }
                        temp.total ++;
                    }
                }
                temp.ave  = temp.tot * 1.0 / temp.total;
                temp.area /= temp.total;
                q[count++] = temp;
            }
        }
        sort (q, q + count, comp);
        printf("%d
    ", count);
        for (int i = 0; i < count; i++)
        {
            printf("%04d %d %.3lf %.3lf
    ", q[i].id, q[i].total, q[i].ave, q[i].area);
        }
    }
    
    
    
    
    
    
    
    
    
    
    
    
    
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  • 原文地址:https://www.cnblogs.com/Dup4/p/9433235.html
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